原文链接: 1094. The Largest Generation (25)
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1094. The Largest Generation (25)
https://www.patest.cn/contests/pat-a-practise/1094
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.
Input Specification:
Each input file contains one test case. Each case starts with two positive integers N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:
ID K ID[1] ID[2] ... ID[K]
where ID is a two-digit number representing a family member, K (>0) is the number of his/her children, followed by a sequence of two-digit ID's of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.
Sample Input:
23 13 21 1 23 01 4 03 02 04 05 03 3 06 07 08 06 2 12 13 13 1 21 08 2 15 16 02 2 09 10 11 2 19 20 17 1 22 05 1 11 07 1 14 09 1 17 10 1 18
Sample Output:
9 4
题目大意:输入树的结点个数N,结点编号为1~N,非叶子结点个数M,然后输出M个非叶子结点格子的孩子结点的编号,求结点个数最多的一层,根结点的层号为1,输出该层的结点个数以及层号<(▰˘◡˘▰)>
分析:用DFS或者BFS,用DFS就用参数index和level标记当前遍历的结点的编号和层数,一个数组book标记当前层数level所含结点数,最后遍历一遍数组找出最大值。注意 :book[level]++;这句话是发生在return语句判断之前的外面,即每遇到一个结点都要进行处理,而不是放在return语句的条件判断里面~~~~~
如果是BFS,就用一个数组level[i]标记i结点所处的层数,它等于它的父亲结点的level的值+1,用一个数组book,book[i]标记i层所拥有的结点数,在遍历的时候每弹出一个结点就将当前结点的层数所对应的book值+1,最后遍历一遍book数组找出最大拥有的结点数和层数~~~
#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
vector< vector<int> > tree;
int num[111] = { 0 }; //num[i] 表示存放第i代的人数
int maxDepth = 0; //最大代数
//树根节点是第一代
void dfs(int root, int depth) {
num[depth]++;
if (tree[root].empty()) {
maxDepth = maxDepth < depth ? depth : maxDepth;
return;
}
for (int i = 0; i < tree[root].size(); i++)
dfs(tree[root][i], depth + 1);
}
int main(int argc, char *argv[]) {
int n, m;
scanf("%d%d", &n, &m);
tree.resize(n + 1);
for (int i = 0; i < m; i++) {
int f, cnt, k;
scanf("%d %d", &f, &cnt);
for (int j = 0; j < cnt; j++) {
scanf("%d", &k);
tree[f].push_back(k);
}
}
dfs(1, 1); //根节点是1
int ans = 1; //有人数最多的最大代数
for (int i = 2; i <= maxDepth; i++) {
if (num[ans] < num[i])
ans = i;
}
printf("%d %d\n", num[ans], ans);
return 0;
}
#include <cstdio>
#include <vector>
#include <queue>
using namespace std;
vector<int> v[100];
int level[100];
int book[100];
int main() {
int n, m, a, k, c;
scanf("%d %d", &n, &m);
for(int i = 0; i < m; i++) {
scanf("%d %d",&a, &k);
for(int j = 0; j < k; j++) {
scanf("%d", &c);
v[a].push_back(c);
}
}
queue<int> q;
q.push(1);
level[1] = 1;
while(!q.empty()) {
int index = q.front();
q.pop();
book[level[index]]++;
for(int i = 0; i < v[index].size(); i++) {
level[v[index][i]] = level[index] + 1;
q.push(v[index][i]);
}
}
int maxnum = 0, maxlevel = 1;
for(int i = 1; i < 100; i++) {
if(book[i] > maxnum) {
maxnum = book[i];
maxlevel = i;
}
}
printf("%d %d", maxnum, maxlevel);
return 0;
}