Network-POJ3694（最小公共祖先LCA+Tarjin）

http://poj.org/problem?id=3694

 

这一题  为什么要找最小祖先呢

当两个节点连到一块的时候  找最小公共节点就相当于找强连通分支

再找最小公共节点的过程中直到找到  这个过程中所有的点就是一个强连通分支

现在要求桥   只需用没有加边的时候的桥数减去后来找到的强连通分支里的桥数就得到加边后的桥数

 

Network

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7720		Accepted: 2823

Description

A network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers are connected directly or indirectly by successive links, so data can be transformed between any two computers. The administrator finds that some links are vital to the network, because failure of any one of them can cause that data can't be transformed between some computers. He call such a link a bridge. He is planning to add some new links one by one to eliminate all bridges.

You are to help the administrator by reporting the number of bridges in the network after each new link is added.

Input

The input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤ 200,000).
Each of the following M lines contains two integers A and B ( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B. Computers are numbered from 1 to N. It is guaranteed that any two computers are connected in the initial network.
The next line contains a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links the administrator plans to add to the network one by one.
The i-th line of the following Q lines contains two integer A and B (1 ≤ A ≠ B ≤ N), which is the i-th added new link connecting computer A and B.

The last test case is followed by a line containing two zeros.

Output

For each test case, print a line containing the test case number( beginning with 1) and Q lines, the i-th of which contains a integer indicating the number of bridges in the network after the first i new links are added. Print a blank line after the output for each test case.

Sample Input

3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0

Sample Output

Case 1:
1
0

Case 2:
2
0

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>

using namespace std;
#define N 200000

int low[N],dfn[N],n,fa[N],Stack[N],bridge[N];
int Time,top,ans;
vector<vector <int> >G;

void Inn()
{
    G.clear();
    G.resize(n+1);
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(fa,0,sizeof(fa));
    memset(bridge,0,sizeof(bridge));
    memset(Stack,0,sizeof(Stack));
    Time=top=ans=0;
}

void Tarjin(int u,int f)
{
    dfn[u]=low[u]=++Time;
    fa[u]=f;
    int len=G[u].size(),v;
    for(int i=0; i<len; i++)
    {
        v=G[u][i];
        if(!dfn[v])
        {
            Tarjin(v,u);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
            {
                bridge[v]++;
                ans++;
            }
        }
        else if(v!=f)
            low[u]=min(low[u],dfn[v]);
    }
}
void LCA(int a,int b)
{
    if(a==b)
        return;
    if(dfn[a]>dfn[b])
    {
        int v=fa[a];
        if(bridge[a]>0)
        {
            bridge[a]=0;
        ans--;
        }
        LCA(v,b);
    }
    else
    {
        int v=fa[b];
        if(bridge[b]>0)
        {
            bridge[b]=0;
        ans--;
        }
        LCA(a,v);
    }
}

int main()
{
    int m,a,b,q,i,t=1;
    while(scanf("%d %d",&n,&m),n+m)
    {
        Inn();
        for(i=1; i<=m; i++)
        {
            scanf("%d %d",&a,&b);
            G[a].push_back(b);
            G[b].push_back(a);
        }
        Tarjin(1,0);
        scanf("%d",&q);
        printf("Case %d:\n",t++);
        while(q--)
        {
            scanf("%d %d",&a,&b);
            LCA(a,b);
            printf("%d\n",ans);
        }
    }
    return 0;
}