hdu2460&&poj3694 缩点+lca变形

http://acm.hdu.edu.cn/showproblem.php?pid=2460

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

Problem 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

/**hdu2460&&poj3694   缩点+lca变形题目大意：给定一个图，然后依次加一些边，求每加入一条边后现有的图中含有多少桥解题思路：先把所有的强连通分量进行缩点，然后现有桥的个数为点数减一，而后每增加一条边u->v那么u，v到它们lca所在的环之间的桥都要减去。           先dfs将所有的桥标记（采用标记点的方式标记桥），然后每次加边来一次向根节点查找就好了*/#pragma comment(linker, "/STACK:10240000000000,10240000000000")/// 申请空间hdu需要#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>using namespace std;const int maxn=200005;int head[maxn],ip;int m,n,bridge,ans;int low[maxn],dfn[maxn],dex,cnt,st[maxn],inst[maxn],belong[maxn],top;void init(){    memset(head,-1,sizeof(head));    ip=0;}struct note{    int v,cut,next;}edge[maxn*2];void addedge(int u,int v){    edge[ip].v=v,edge[ip].cut=0,edge[ip].next=head[u],head[u]=ip++;}void tarjan(int u,int pre){    dfn[u]=low[u]=++dex;    st[top++]=u;    inst[u]=1;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==pre)continue;        if(dfn[v]==0)        {            tarjan(v,u);            if(low[u]>low[v])low[u]=low[v];            if(low[v]>low[u])            {                bridge++;                edge[i].cut=1;                edge[i^1].cut=1;            }        }        else if(inst[v]&&low[u]>dfn[v])        {            low[u]=dfn[v];        }    }    if(dfn[u]==low[u])    {        int j;        cnt++;        do        {            j=st[--top];            inst[j]=0;            belong[j]=cnt;        }        while(j!=u);    }}vector <int> vec[maxn];int father[maxn];int dep[maxn];int a[maxn];void bfs(int root){    memset(dep,-1,sizeof(dep));    dep[root]=0;    a[root]=0;    father[root]=-1;    queue<int>q;    q.push(root);    while(!q.empty())    {        int tmp=q.front();        q.pop();        for(int i=0;i<vec[tmp].size();i++)        {            int v=vec[tmp][i];            if(dep[v]!=-1)continue;            dep[v]=dep[tmp]+1;            a[v]=1;            father[v]=tmp;            q.push(v);        }    }}void lca(int u,int v){    if(dep[u]>dep[v])swap(u,v);    while(dep[u]<dep[v])    {        if(a[v])        {            ans--;            a[v]=0;        }        v=father[v];    }    while(u!=v)    {        if(a[u])        {            ans--;            a[u]=0;        }        if(a[v])        {            ans--;            a[v]=0;        }        v=father[v];        u=father[u];    }}void solve(){    memset(dfn,0,sizeof(dfn));    memset(inst,0,sizeof(inst));    cnt=dex=top=bridge=0;    tarjan(1,-1);    for(int i=0;i<n;i++)        vec[i].clear();    for(int u=1;u<=n;u++)    {        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].v;            if(edge[i].cut)            {                int x=belong[u];                int y=belong[v];                vec[x].push_back(y);                vec[y].push_back(x);            }        }    }    bfs(1);    int Q;    //printf("%d %d\n",cnt-1,bridge);    ans=cnt-1;    scanf("%d",&Q);    while(Q--)    {        int u,v;        scanf("%d%d",&u,&v);        lca(belong[u],belong[v]);        printf("%d\n",ans);    }    printf("\n");}int main(){    int T,tt=0;    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0)break;        init();        for(int i=0;i<m;i++)        {            int u,v;            scanf("%d%d",&u,&v);            addedge(u,v);            addedge(v,u);        }        printf("Case %d:\n",++tt);        solve();    }    return 0;}