LCA+桥poj3694

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Network

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 6730 Accepted: 2389

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 21 22 321 21 34 41 22 12 31 421 23 40 0

Sample Output

Case 1:10Case 2:20

先找出图中的桥，然后并查集缩点，询问的时候，找出连点的LCA，然后把经过的连通分量再合并

#include<iostream>#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int maxn=100010;int n,m,q;int pre[maxn],low[maxn],head[maxn],dfs_clock,fa[maxn],cnt;vector<int> g[maxn];int find(int x){    if(x==head[x])return x;    return head[x]=find(head[x]);}void unite(int a,int b){    int x=find(a),y=find(b);    head[y]=x;}void dfs(int u,int f){    low[u]=pre[u]=++dfs_clock;    int len=g[u].size();    for(int i=0;i<len;i++)    {        int v=g[u][i];        if(!pre[v])        {            fa[v]=u;            dfs(v,u);            low[u]=min(low[u],low[v]);            if(low[v]>pre[u])cnt++;            else unite(u,v);        }        else if(pre[v]<pre[u]&&v!=f)            low[u]=min(low[u],pre[v]);    }}void find_bcc(){    memset(pre,0,sizeof(pre));    for(int i=0;i<=n;i++)head[i]=i;    dfs_clock=cnt=0;    dfs(1,-1);}void LCA(int a,int b){    while(a!=b)    {        while(pre[a]>=pre[b]&&a!=b)        {            if(find(a)!=find(fa[a]))            {                cnt--;                unite(fa[a],a);            }            a=fa[a];        }        while(pre[b]>=pre[a]&&a!=b)        {            if(find(b)!=find(fa[b]))            {                cnt--;                unite(fa[b],b);            }            b=fa[b];        }    }    printf("%d\n",cnt);}int main(){    int cas=1;    while(scanf("%d%d",&n,&m)!=EOF,n||m)    {        for(int i=1;i<=m;i++)        {            int u,v;            scanf("%d%d",&u,&v);            g[u].push_back(v);            g[v].push_back(u);        }        find_bcc();        scanf("%d",&q);        printf("Case %d:\n",cas++);        while(q--)        {            int l,r;            scanf("%d%d",&l,&r);            if(find(l)==find(r))printf("%d\n",cnt);            else LCA(l,r);        }        printf("\n");    }    return 0;}