CSU-1908 The Big Escape

CSU-1908 The Big Escape

Description

There is a tree-like prison. Expect the root node, each node has a prisoner, and the root node is the only exit. Each node can accommodate a large number of prisoners, but each edge per minute only one prisoner can pass.
Now, the big escape begins, every prisoner wants to escape to the exit.Do you know when the last one escapes from the prison.

Input

There are lots of case.
For each test case.The first line contains two integers n,m（n<=100000, 1<=m<=n）, which indicates the number of nodes and the root node.
The next n-1 lines describe the tree.

Output

For each test case, you output one line “Case #%d:%d”

Sample Input

10 2
1 2
2 3
2 4
2 5
1 6
5 7
3 8
2 9
2 10

Sample Output

Case #1:2

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题解

题意是给定一个树形监狱，每条边每分钟只能允许一个人通过，给定根节点，犯人逃到根节点就算逃出，每个节点可以存在多个犯人，问逃出的最短时间

这个题就是统计根节点最大子树有多少节点

#include<bits/stdc++.h>
#define maxn 100050
using namespace std;
vector<int> G[maxn];
int sum;
void dfs(int u, int fa) {
	for (int i = 0; i < G[u].size(); i++) {
		int v = G[u][i];
		if (v == fa) continue;
		sum++;
		dfs(v, u);
	}
}
int main() {
	int n, m;
	int cnt = 0;
	while (scanf("%d%d", &n, &m) != EOF) {
		cnt++;
		for (int i = 1; i <= 100000; i++) {
			G[i].clear();
		}
		for (int i = 1; i < n; i++) {
			int a, b;
			scanf("%d%d", &a, &b);
			G[a].push_back(b);
			G[b].push_back(a);
		}
		int ans = 0;
		for (int i = 0; i < G[m].size(); i++) {
			int v = G[m][i];
			sum = 0;
			dfs(v, m);
			ans = max(ans, sum);
		}
		printf("Case #%d:%d
", cnt, ans + 1);
	}
}
/**********************************************************************
	Problem: 1908
	User: Artoriax
	Language: C++
	Result: AC
	Time:748 ms
	Memory:8064 kb
**********************************************************************/