hdu 5452 Minimum Cut 树形dp

Minimum Cut

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5452

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

HINT

题意

给你一个没有自环重边的无向图，然后给你一颗生成树，让你求一个最小割集的大小，使得这个割集有且只有一条边在这棵生成树上

题解：

树形dp，对于这个点，如果要消除他和他父亲之间的联系的话，代价就是他的子树所有连向外界的边就好了

跑一遍dfs维护一下就行了

-------------------------

我的方法是错的，已经被hack了

数据：

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

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 20050
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
//*********************************************************************************

vector<int> Q[maxn];
vector<int> E[maxn];
int vis[maxn];
int dp[maxn];
int ans;
void dfs(int x)
{
    vis[x]=1;
    for(int i=0;i<Q[x].size();i++)
    {
        int y=Q[x][i];
        dfs(Q[x][i]);
        dp[x]+=dp[y]-1;
    }
    ans = min(ans,dp[x]);
    for(int i=0;i<E[x].size();i++)
        if(vis[E[x][i]])
            dp[E[x][i]]--;
    vis[x]=0;
}
int main()
{
    int t;scanf("%d",&t);
    for(int cas = 1;cas<=t;cas++)
    {
        ans = 99999999;
        int n,m;scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++)
            Q[i].clear(),E[i].clear();
        memset(dp,0,sizeof(dp));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<n-1;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            dp[x]++,dp[y]++;
            if(x>y)swap(x,y);
            Q[x].push_back(y);
        }
        for(int i=n-1;i<m;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            dp[x]++,dp[y]++;
            if(x>y)swap(x,y);
            E[y].push_back(x);
        }
        dfs(1);
        printf("Case #%d: %d
",cas,ans);
    }
}