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HZNU第十二届校赛赛后补题

o
 osc_fmg49rzg
发布于 2019/03/20 15:24
字数 5144
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愉快的校赛翻皮水!

 

题解

A 温暖的签到,注意用gets

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1000000 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
string a; 
int main(){
    while(getline(cin,a)){
        a.back() = '!';
        cout << a << endl;
    }
    return 0;
}
A

 

B.比赛的时候一直以为是主席树上操作或者其他的高级数据结构,万万没想到是在序列特性下手,打一张最小的不冲突的表就会发现斐波那契数列是最小不冲突序列,int范围内最多容纳47个数左右,所以小于50的范围暴力查询,大于50的范围必定YES

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
LL a[maxn];
LL b[maxn];
int main(){
    Sca2(N,M);
    for(int i = 1; i <= N ; i ++) Scl(a[i]);
    for(int i = 1; i <= M ; i ++){
        int l,r; Sca2(l,r);
        if(r - l + 1 < 3){
            puts("NO");
            continue;
        }
        if(r - l + 1 >= 50) puts("YES");
        else{
            int flag = 0;
            for(int j = l; j <= r; j ++) b[j] = a[j];
            sort(b + l,b + r + 1);
            for(int j = l + 2; j <= r; j ++){
                if(b[j - 2] + b[j - 1] > b[j]){
                    flag = 1;
                    break;
                }
            }
            if(flag) puts("YES");
            else puts("NO");
        }
    }
    return 0; //1 1 2 3 5 8 13 21 34
}
B

 

D. 8说了,温暖的签到

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int a[maxn];
set<int>Q;
int main(){
    N = read();
    while(N--) Q.insert(read());
    Pri(Q.size());
    return 0; //1 1 2 3 5 8 13 21 34
}
D

 

E.给一个条件构造的图,求图上的哈密顿回路。

可以猜想到N为奇数的时候始终不可行。

N为偶数的时候必定可行,并且可以发现,每个点都恰好有两个出度和两个入度,

这就可以转换成欧拉回路直接做

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 10010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int ans[maxn],cnt;
bool vis[maxn];
void dfs(int t){
    int l = (t * 2) % N,r = (t * 2 + 1) % N;
    if(l > r) swap(l,r);
    if(!vis[r]){vis[r] = 1;dfs(r);}
    if(!vis[l]){vis[l] = 1;dfs(l);}
    ans[++cnt] = t;
}
int main(){
    Sca(N);
    if(N & 1){puts("-1"); return 0;}
    dfs(0);
    for(int i = cnt ; i >= 1; i --)cout << ans[i] << " " ;
    return 0;
}
E

 

G.由于每天加的钱为实数而不要求为浮点数,一个显然的贪心是每两个取的物品i,j之间,每天加入的钱都是wj / (j - i)

dp是显然的,第一个难点在于每天钱数不增的限制,如果dp存储加入的钱数,2000 * 1e6很显然时间复杂度上过不去

一个比较巧妙地思想是dp[i][j]表示上一个操作是i - > j的物品的选择,2000 * 2000满足了时间复杂度还满足了钱数

得到状态转移方程dp[j][k] = max(dp[i][j] + V[k])

到了这一步就可以写出一个n3 的暴力(雾),将状态转移方程变形,得到i > j - W(j) / W(k) * (k - j)这样一个右边和i无关的方程.

所以考虑枚举j和k,然后就可以得到i的下界,i的上界显然为j - 1,就变成了一个后缀最大值的问题,这个甚至不需要树状数组,直接维护一个简单的一维数组即可.

时间复杂度n²

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-6;
const int maxn = 2010;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N;
struct Good{
    LL v;
    double w;
}P[maxn];
LL dp[maxn][maxn];
LL Max[maxn];
int main(){
    Sca(N);
    for(int i = 1; i <= N ; i ++) scanf("%lf",&P[i].w);
    for(int i = 1; i <= N ; i ++) Scl(P[i].v);
    for(int i = 0; i <= N ; i ++){
        for(int j = 0; j <= N ; j ++){
            dp[i][j] = -1e18;
        }
    }
    for(int i = 1; i <= N ; i ++) dp[0][i] = P[i].v;
    for(int j = 1; j <= N ; j ++){
        Max[j] = -1e18;
        for(int k = j - 1; k >= 0; k --) Max[k] = max(Max[k + 1],dp[k][j]);
        for(int k = j + 1; k <= N ; k ++){
            double l;
            if(!P[k].w) l = 0;
            else l = max(j - P[j].w / P[k].w * (k - j),(double)0);
            int L = (int)l;
            if(fabs(l - L) > eps) L++;
            if(L > j - 1) continue;
            dp[j][k] = Max[L] + P[k].v; // Max[k][j]
        }
    }
    LL ans = 0;
    for(int i = 0; i <= N ; i ++){
        for(int j = i + 1; j <= N ; j ++){
            ans = max(ans,dp[i][j]);
        }
    }
    Prl(ans);
    return 0;
}
G

 

H.取个log就会发现变成AjlogAi > AilogAj,直接sort一波,特判一下前后相等的情况即可.

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
PIL a[maxn];
int ans[maxn];
bool cmp(PIL x,PIL y){
    return 1.0 * x.se * log(y.se * 1.0) > 1.0 * y.se * log(x.se * 1.0);
}
bool equal(double x,double y){
    return fabs(x - y) < eps;
}
int main(){
    N = read();
    for(int i = 1; i <= N ; i ++){
        Scl(a[i].se);
        a[i].fi = i;
    } 
    sort(a + 1,a + 1 + N,cmp);
    int cnt = 0;
    for(int i = 1; i <= N ; i ++){
        if(i > 1 && equal(1.0 * a[i].se * log(a[i - 1].se * 1.0),1.0 * a[i - 1].se * log(a[i].se * 1.0))) cnt++;
        else cnt = 0;
        ans[a[i].fi] = i - 1 - cnt;
    }
    for(int i = 1; i <= N ; i ++){
        printf("%d ",ans[i]);
    }
    return 0;
}
H

 

I.要有多坑有多坑,正确题意为双射 + 最后25字母可推26字母,其他没有难度

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
int to[30],to2[30];
char str[maxn],str2[maxn];
int main(){
    while(~scanf("%s%s",str,str2)){
        int l = strlen(str);
        for(int i = 0 ; i < 26; i ++) to[i] = to2[i] = -1;
        for(int i = 0; str[i]; i ++){
            int id1 = str[i] - 'a',id2 = str2[i] - 'a';
            if(~to[id1] && to[id1] != id2){
                puts("Impossible"); exit(0);
            }
            if(~to2[id2] && to2[id2] != id1){
                puts("Impossible"); exit(0);
            }
            to[id1] = id2; to2[id2] = id1;
        }
        int cnt = 0;
        for(int i = 0 ; i < 26; i ++) if(~to[i]) cnt++;
        if(cnt == 25){
            int t = 0;
            for(int i = 0 ; i < 26; i ++){
                if(to[i] == -1) t = i;
            }
            for(int i = 0 ; i < 26; i ++) if(to2[i] == -1) to[t] = i;
        }
        for(int i = 0 ; i < 26; i ++){
            if(~to[i]) printf("%c->%c\n",i + 'a',to[i] + 'a');
        }
    }
    return 0;
}
I

 

J.过于温暖,给出题人点赞

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 110;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
LL N,M,K,T;
int main(){
    cin >> N >> K >> T;
    cout << max(0LL,N - K * T) << endl;
    return 0;
}
J

 

K.很显然是预处理出所有情况然后二分端点.

坑点1.不能用海伦公式,过不了double浮点数的误差,需要用向量叉积并且不除2,后面查询的时候将l和r乘2达到无浮点数的目的

2.r用upper_bound,l用lower_bound,最后r - l即可,天知道为什么我一开始手写了两个二分

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 310;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,Q;
struct Point{
    LL x,y;
}point[maxn];
LL P[255 * 255 * 50];
LL cul(LL x1,LL y1,LL x2,LL y2){
    //cout << x1 << " " << y1 << " " << x2 << " " << y2 << endl;
    return abs(x1 * y2 - x2 * y1);
}
int main(){
    Sca2(N,Q);  //x1:(ax - bx) y1:(ay - by) x2:ax - cx y2:ay - cy
    for(int i = 1; i <= N ; i ++) scanf("%lld%lld",&point[i].x,&point[i].y);
    int cnt = 0;
    for(int i = 1; i <= N ; i ++){
        for(int j = i + 1; j <= N; j ++){
            LL x1 = point[i].x - point[j].x,y1 = point[i].y - point[j].y;
            for(int k = j + 1; k <= N ; k ++){
                LL x2 = point[i].x - point[k].x,y2 = point[i].y - point[k].y;
                P[++cnt] = cul(x1,y1,x2,y2);
            }
        }
    }
    sort(P + 1,P + 1 + cnt);
    //for(int i = 1; i <= cnt; i ++) cout << P[i] << " ";
    //cout << endl;
    for(int i = 1; i <= Q; i ++){
        LL l,r; scanf("%lld%lld",&l,&r);
        l *= 2; r *= 2;
        l = lower_bound(P + 1,P + 1 + cnt,l) - P;
        r = upper_bound(P + 1,P + 1 + cnt,r) - P;
        //cout << l << ' ' << r << endl;
        Pri(r - l);
    }
    return 0;
}
K

 

L.考虑从T开始跑一颗最短路树,那么如果当人在i点的时候去掉边,显然去掉非树边上的边的行为是无意义的,如果去掉了树边上的边,那么就需要走到他们的子树上某个结点j,通过一个非树边走到一个非子树的结点k,然后再到T点,距离就是dis[j] - dis[i] + edge(j,k) + dis[k]

显然(雾)我们可以对每一条非树边进行预处理,每一条连接(i,j)的非树边将会对i到lca(i,j),j到lca(i,j)的两条链产生贡献(目的是寻找一条i点树边被封之后最短的走到T的路线),这个过程可以用树链剖分实现

最后再从T开始往S跑最短路,由于我们已经求出了每个点i的树边被封之后到T的最短路,所以更新答案是dis[v] = max(dis[u.pos] + edge[i].dis,dis2[v]);

也就是说,如果在v点直接切断会比在之后切断造成需要走更长的路。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d\n", x)
#define Prl(x) printf("%lld\n",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}
while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f;}
const double eps = 1e-9;
const int maxn = 2e5 + 10;
const int maxm = 2e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,S,T;
struct Edge{
    int to,next;
    LL dis;
}edge[maxm * 2];
struct E{
    int u,v,flag;
    LL w;
    E(){}
    E(int u,int v,LL w):u(u),v(v),w(w){}
}e[maxm * 2];
int head[maxn],tot; 
void init(){
    for(int i = 0 ; i <= N ; i ++) head[i] = -1;
    tot = 0;
}
void add(int u,int v,LL w){
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].dis = w;
    head[u] = tot++;
}
struct node{
    int pos;
    LL cost;
    node(){}
    node(int pos,LL cost):pos(pos),cost(cost){}
    friend bool operator < (node a,node b){
        return a.cost > b.cost;
    }
};
LL dis[maxn],dis2[maxn];
int fa[maxn];
void Dijkstra(int s){
    for(int i = 1; i <= N ; i ++) dis[i] = 1e18;
    priority_queue<node>Q;
    Q.push(node(s,0)); fa[s] = dis[s] = 0;
    while(!Q.empty()){
        node u = Q.top(); Q.pop();
        if(u.cost > dis[u.pos]) continue; 
        for(int i = head[u.pos]; ~i; i = edge[i].next){
            int v = edge[i].to;
            if(dis[v] > dis[u.pos] + edge[i].dis){
                dis[v] = dis[u.pos] + edge[i].dis;
                fa[v] = u.pos;
                Q.push(node(v,dis[v]));
            }
        }
    }
}
int dep[maxn],top[maxn],pos[maxn],size[maxn],son[maxn];
void dfs1(int t,int la){
    size[t] = 1; son[t] = t;
    int heavy = 0;
    for(int i = head[t]; ~i ; i = edge[i].next){
        int v = edge[i].to;
        if(v == la) continue;
        dep[v] = dep[t] + 1;
        fa[v] = t;
        dfs1(v,t);
        if(size[v] > heavy){
            heavy = size[v];
            son[t] = v;
        }
        size[t] += size[v];
    }
}
int cnt;
void dfs2(int t,int la){
    top[t] = la;
    pos[t] = ++cnt;
    if(son[t] == t) return;
    dfs2(son[t],la);
    for(int i = head[t]; ~i ; i = edge[i].next){
        int v = edge[i].to;
        if((fa[t] == v) || v == son[t]) continue;
        dfs2(v,v);
    }
}
int lca(int u,int v){
    while(top[u] != top[v]){
        if(dep[top[u]] < dep[top[v]]) swap(u,v);
        u = fa[top[u]];
    }
    if(dep[u] < dep[v]) return u;
    return v;
}
struct Tree{
    int l,r;
    LL lazy;
}tree[maxn << 2];
void Build(int t,int l,int r){
    tree[t].l = l; tree[t].r = r;
    tree[t].lazy = 1e18;
    if(l == r) return;
    int m = l + r >> 1;
    Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r);
}
void Pushdown(int t){
    tree[t << 1].lazy = min(tree[t << 1].lazy,tree[t].lazy);
    tree[t << 1 | 1].lazy = min(tree[t << 1 | 1].lazy,tree[t].lazy);
}
void update(int t,int l,int r,LL sum){
    if(l <= tree[t].l && tree[t].r <= r){
        tree[t].lazy = min(tree[t].lazy,sum);
        return ;
    }
    Pushdown(t);
    int m = tree[t].l + tree[t].r >> 1;
    if(r <= m) update(t << 1,l,r,sum);
    else if(l > m) update(t << 1 | 1,l,r,sum);
    else{
        update(t << 1,l,m,sum);
        update(t << 1 | 1,m + 1,r,sum);
    }
}
LL query(int t,int p){
    if(tree[t].l == tree[t].r) return tree[t].lazy; 
    Pushdown(t);
    int m = tree[t].l + tree[t].r >> 1;
    if(p <= m) return query(t << 1,p);
    else return query(t << 1 | 1,p);
}
void update(int u,int v,LL sum){
    while(top[u] != top[v]){
        update(1,pos[top[u]],pos[u],sum);
        u = fa[top[u]];
    }
    if(u == v) return;
    update(1,pos[v] + 1,pos[u],sum);
}
int main(){
    scanf("%d%d%d%d",&N,&M,&S,&T); init();
    for(int i = 1; i <= M ; i ++){
        int u,v; Sca2(u,v); LL w; Scl(w);
        add(u,v,w); add(v,u,w);
        e[i] = E(u,v,w); e[i].flag = 1;
    }
    Dijkstra(T); init();
    for(int i = 1; i <= M ; i ++){
        if(fa[e[i].u] == e[i].v || fa[e[i].v] == e[i].u){
            add(e[i].v,e[i].u,e[i].w);
            add(e[i].u,e[i].v,e[i].w);
        }else e[i].flag = 0;
    }
    dfs1(T,T); dfs2(T,T);
    Build(1,1,N);
    for(int i = 1; i <= M ; i ++){
        if(e[i].flag) continue;
        LL sum = dis[e[i].u] + dis[e[i].v] + e[i].w;
        int l = lca(e[i].u,e[i].v);
        update(e[i].u,l,sum); update(e[i].v,l,sum);
    }
    for(int i = 1; i <= N; i ++) dis2[i] = query(1,pos[i]) - dis[i];
    for(int i = 1; i <= N ; i ++) dis[i] = 1e18;  init();
    for(int i = 1; i <= M ; i ++){
        add(e[i].u,e[i].v,e[i].w);
        add(e[i].v,e[i].u,e[i].w);
    } 
    priority_queue<node>Q;
    Q.push(node(T,0)); dis[T] = 0;
    while(!Q.empty()){
        node u = Q.top(); Q.pop();
        if(dis[u.pos] < u.cost) continue;
        for(int i = head[u.pos]; ~i ; i = edge[i].next){
            int v = edge[i].to;
            if(dis[v] > max(dis[u.pos] + edge[i].dis,dis2[v])){
                dis[v] = max(dis[u.pos] + edge[i].dis,dis2[v]);
                Q.push(node(v,dis[v]));
            }
        }
    }
    if(dis[S] >= 1e16) puts("-1");
    else Prl(dis[S]);
    return 0;
}
L

 

M.用SG函数打表博弈,反正我是不会做,贴上队友代码

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <set>
#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>


using namespace std;

typedef long long LL;
const int maxn = 1e6+500;
int if_pre[maxn +5];

void pre_first()
{
    memset(if_pre, 0x3f, sizeof(if_pre));
    if_pre[0] = 0;
    if_pre[1] = 1;
    int pre_num =2;
    for (int i = 2; i < maxn; i++)
    {
        if (if_pre[i] == 0x3f3f3f3f)
        {
            if_pre[i] = pre_num++;
            for (int j = i << 1; j < maxn; j += i)
            {
                if_pre[j] = min(if_pre[j],if_pre[i]);
            }
        }
    }
}

int main()
{
    int t;
    cin >>t;
    pre_first();
    while (t --)
    {
        int n,a;
        cin >>n;
        int sum =0;
        while (n--)
        {
            cin >>a;
            sum^=if_pre[a];
            //cout<<a<<'*'<<if_pre[a]<<endl;
        }
        if (sum)
            cout<<"Subconscious is our king!"<<endl;
        else
            cout<<"Long live with King Johann!"<<endl;
        
    }




#ifdef VSCode
    system("pause");
#endif
    return 0;
}
M

 

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