## 1146. Topological Order (25) 转

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osc_z1hvg4cu

This is a problem given in the Graduate Entrance Exam in 2018: Which of the following is NOT a topological order obtained from the given directed graph? Now you are supposed to write a program to test each of the options. Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers N (<= 1,000), the number of vertices in the graph, and M (<= 10,000), the number of directed edges. Then M lines follow, each gives the start and the end vertices of an edge. The vertices are numbered from 1 to N. After the graph, there is another positive integer K (<= 100). Then K lines of query follow, each gives a permutation of all the vertices. All the numbers in a line are separated by a space.

Output Specification:

Print in a line all the indices of queries which correspond to "NOT a topological order". The indices start from zero. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line. It is graranteed that there is at least one answer.

Sample Input:
``````6 8
1 2
1 3
5 2
5 4
2 3
2 6
3 4
6 4
5
1 5 2 3 6 4
5 1 2 6 3 4
5 1 2 3 6 4
5 2 1 6 3 4
1 2 3 4 5 6``````
Sample Output:
``````3 4

``````#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#define Max 1005
using namespace std;
int n,m,k,limit[Max],exa[Max],a,b,mp[Max][Max],ans[Max],ant;///limit存某个点前面有几个点先行

int check()
{
int p[Max];
for(int i = 1;i <= n;i ++)///limit转到p
{
p[i] = limit[i];
}
for(int i = 0;i < n;i ++)
{
if(p[exa[i]])///p为正表示顺序不合法
{
return 0;
}
for(int j = 1;j <= n;j ++)///如果合法 把受它限制的点 p都减1
{
if(mp[exa[i]][j])p[j] --;
}
}
return 1;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i = 0;i < m;i ++)
{
scanf("%d%d",&a,&b);
mp[a][b] = 1;
limit[b] ++;
}
scanf("%d",&k);
for(int i = 0;i < k;i ++)
{
for(int j = 0;j < n;j ++)
{
scanf("%d",&exa[j]);
}
if(!check())
{
ans[ant ++] = i;
}
}
for(int i = 0;i < ant;i ++)
{
if(i)printf(" %d",ans[i]);
else printf("%d",ans[i]);
}
}``````

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