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Time Limit: 1 Sec
Memory Limit: 256 MB
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.
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.
For each test case, you should output the minimum cut of graph G respecting the given spanning tree T.
Sample Output
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
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\n",cas,ans); } }
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原文地址：http://www.cnblogs.com/qscqesze/p/4822468.html
