HDU5452 Minimum Cut（图论）

Minimum Cut

传送门1
传送门2
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

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题意

给出一个图$G$，求删除最少的边使的图$G$变为不连通，所删除的边有且仅有一条属于图$G$的生成树$T$.

分析

显然我们要把一个点从图中分离。那么这个点在树上的度一定为1，再取图上另外要删除的边最少的点，它的度就是另外要删除的边的条数，最后答案加1.

CODE

#include<cstdio>
#include<memory.h>
#define N 20005
int ct[N],dg[N];
int main() {
    int T;
    scanf("%d",&T);
    for(int cas=1; cas<=T; cas++) {
        int n,m;
        scanf("%d%d",&n,&m);
        memset(ct,0,sizeof ct);
        memset(dg,0,sizeof dg);
        int u,v;
        for(int i=1; i<n; i++) {
            scanf("%d%d",&u,&v);
            dg[u]++;
            dg[v]++;
        }
        for(int i=n; i<=m; i++) {
            scanf("%d%d",&u,&v);
            ct[u]++;
            ct[v]++;
        }
        int res=2e9;
        for(int i=2; i<=n; i++)if(dg[i]==1)if(res>ct[i])res=ct[i];
        printf("Case #%d: %d\n",cas,res+1);
    }
    return 0;
}