hdu2460&&poj3694 縮點+lca變形，

hdu2460&&poj3694 縮點+lca變形，

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

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

Problem DescriptionA 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. 
InputThe 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. 
OutputFor 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   縮點+lca變形題目大意：給定一個圖，然後依次加一些邊，求每加入一條邊後現有的圖中含有多少橋解題思路：先把所有的強連通分量進行縮點，然後現有橋的個數為點數減一，而後每增加一條邊u->v那麼u，v到它們lca所在的環之間的橋都要減去。           先dfs將所有的橋標記（採用標記點的方式標記橋），然後每次加邊來一次向根節點尋找就好了*/#pragma comment(linker, "/STACK:10240000000000,10240000000000")/// 申請空間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;}