hdu5452 || 瀋陽網路賽1003題 最近公共祖先問題

life4711發表於2015-09-22

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

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


/*
hdu5452 || 瀋陽網路賽1003題 最近公共祖先問題
題目大意：給定一個圖的一棵生成樹然後給出一些其他的邊，沒有重邊和自環，問在取且僅取一條樹邊的前提下，圖的最小割邊的數量是多少
解題思路：num維護每個子樹與外界的相連邊的數量，對於每個非樹枝邊，每加一條其最近公共祖先的num值就要-2，最後用棧統計一下就可以了
*/
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <stack>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=111111;
const int maxm=411111;
int n,m,num[maxn],du[maxn],fa[maxn];

struct note
{
    int v;
    int w;
    int next;
};

struct SGRAPH
{
    int head[maxn],ip;
    note edge[maxm];
    void init()
    {
        memset(head,-1,sizeof(head));
        ip=0;
    }
    void addedge(int u,int v,int c)
    {
        edge[ip].v=v,edge[ip].w=c,edge[ip].next=head[u],head[u]=ip++;
    }
    int d[maxn][20];
    void makeRmqIndex(int A[],int n){
        for(int i=1;i<=n;i++) d[i][0]=i;
        for(int j=1;(1<<j)<=n;j++)
            for(int i=1;i+(1<<j)-1<=n;i++)
                d[i][j] = A[d[i][j-1]] < A[d[i+(1<<(j-1))][j-1]]? d[i][j-1]:d[i+(1<<(j-1))][j-1];
    }
    int rmqIndex(int L,int R,int A[])
    {
        int k=0;
        while ((1<<(k+1))<=R-L+1) k++;
        return A[d[L][k]]<A[d[R-(1<<k)+1][k]]? d[L][k]:d[R-(1<<k)+1][k];
    }
    //---------------------
    int E[maxn*2],R[maxn],D[maxn*2],mn;
    void dfs(int u,int p,int d){
        E[++mn]=u;
        D[mn]=d;
        R[u]=mn;
        fa[u]=p;
        for (int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            if (v==p) continue;
            du[u]++;
            dfs(v,u,d+1);
            E[++mn]=u;
            D[mn]=d;
        }
    }
    void LCA_init(){
        mn=0;
        memset(R,0,sizeof(R));
        dfs(1,-1,1);
        makeRmqIndex(D,mn);
        //getd(1,-1,0);
    }
    int LCA(int u,int v){
        if (R[u]>=R[v]) return E[rmqIndex(R[v],R[u],D)];
        else return E[rmqIndex(R[u],R[v],D)];

    }
} solver;

int main()
{
    int T,tt=0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        solver.init();
        for(int i=1;i<n;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            solver.addedge(u,v,1);
            solver.addedge(v,u,1);
        }
        memset(du,0,sizeof(du));
        solver.LCA_init();
        memset(num,0,sizeof(num));
        for(int i=n;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            num[u]++;
            num[v]++;
            int ans=solver.LCA(u,v);
            num[ans]-=2;
        }
        stack<int>st;
        for(int i=1;i<=n;i++)
        {
            if(du[i]==0)st.push(i);
        }
        while(!st.empty())
        {
            int v=st.top();
            st.pop();
            int u=fa[v];
            du[u]--;
            num[u]+=num[v];
            if(du[u]==0&&u>0)
                st.push(u);
        }
        int ans=inf;
        for(int i=2;i<=n;i++)
        {
            //printf("%d::%d\n",i,num[i]);
            ans=min(ans,num[i]);
        }
        printf("Case #%d: %d\n",++tt,ans+1);
    }
    return 0;
}









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