Hdu 5452 Minimum Cut (图论问题) 2015 ACM-ICPC沈阳网赛

 Hdu 5452 Minimum Cut (图论问题) 2015 ACM-ICPC沈阳网赛

 

Minimum Cut

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 1314    Accepted Submission(s): 608

Problem Description

Given a simple unweighted graph  G (an undirected graph containing no loops nor multiple edges) with  n nodes and  m edges. Let  T be a spanning tree of  G.
We say that a cut in  G respects  T if it cuts just one edges of  T.

Since love needs good faith and hypocrisy return for only grief, you should find the minimum cut of graph  G respecting the given spanning tree  T.

 

Input

The input contains several test cases.
The first line of the input is a single integer  t (1≤t≤5) which is the number of test cases.
Then  t test cases follow.

Each test case contains several lines.
The first line contains two integers  n (2≤n≤20000) and  m (n−1≤m≤200000).
The following  n−1 lines describe the spanning tree  T and each of them contains two integers  u and  v corresponding to an edge.
Next  m−n+1 lines describe the undirected graph  G and each of them contains two integers  u and  v corresponding to an edge which is not in the spanning tree  T.

 

Output

For each test case, you should output the minimum cut of graph  G respecting the given spanning tree  T.

 

Sample Input

1 4 5 1 2 2 3 3 4 1 3 1 4

 

Sample Output

Case #1: 2

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online

 

题意：给一个无权有向图G，图上给一颗生成树T，去掉最少的边（必须有一条边在树上）使图不再连通。
分析：生成树T性质，包含图上所有的结点，n即结点数。而选择生成树的叶子结点的边切割不会比非叶子结点更差，因为叶子结点没有子树，所以算法十分明显，

       枚举所有的叶子结点的非生成树边的入度（等于总度数），找到最小的再加一即是答案（必须有一条边在树上）。

 

/*	
	题意：给一个无权有向图G，图上给一颗生成树T，去掉最少的边（必须有一条边在树上）使图不再连通。
	分析：生成树T性质，包含图上所有的结点，n即结点数。而选择生成树的叶子结点的边切割不会比非叶子结点更差，因为叶子结点没有子树，所以算法十分明显，

枚举所有的叶子结点的非生成树边的入度（等于总度数），找到最小的再加一即是答案（必须有一条边在树上）。
*/
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 1<<29
#define eps 1e-9
#define LD long double
#define LL long long
const int maxn = 200000 + 10;
using namespace std;
int rdegree[maxn], vdegree[maxn];
int main()
{
#ifdef LOCAL
	freopen("data.txt", "r", stdin);
#endif
	int T;
	scanf("%d",&T);
	for (int kase = 1; kase <= T; ++kase)
	{
		memset(rdegree, 0, sizeof(rdegree));
		memset(vdegree, 0, sizeof(vdegree));
		int n, m,u, v;
		scanf("%d%d", &n,&m);
		for (int i = 1; i < n; i++)
		{
			
			scanf("%d%d", &u, &v);
			rdegree[u]++;
			rdegree[v]++;
		}
		for (int i = n; i <= m; i++){
			scanf("%d%d", &u, &v);
			vdegree[u]++;
			vdegree[v]++;
		}
		int MinCut = 1 << 30;
		for (int i = 1; i <= n; i++){
			if (rdegree[i]<2){
				MinCut = min(MinCut, vdegree[i]);
			}
		}
		printf("Case #%d: %d\n", kase, MinCut + 1);
	}	
	return 0;
}