poj3694 Network 无向图tarjan求桥+LCA

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Network

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 6617 Accepted: 2341

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

Source

2008 Asia Hefei Regional Contest Online by USTC

题意：给定n个点m条边构成的无向图，有k次询问添加(u,v)边问桥的个数。

思路：先tarjan求出桥的个数以及哪些边为桥，然后对于每次询问(u,v),u-> lca(u,v) v->lca(u,v)的路径中如果之前是桥，那么添加(u,v)之后就不是桥了，程序中bri[i]=0,bri_cnt--，详见代码：

（注意：MAXN设大点，不然RE）

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=100000+100;struct Edge{    int v;    int next;}edge[MAXN<<2];int n,m,edge_cnt,dfs_clock,top,bri_cnt,k;int head[MAXN],dfn[MAXN],low[MAXN],bri[MAXN<<2],fa[MAXN],ins[MAXN],s[MAXN];void init(){    edge_cnt=dfs_clock=top=bri_cnt=0;    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(ins,0,sizeof(ins));    memset(bri,0,sizeof(bri));    for(int i=1;i<=n;i++)        fa[i]=i;}void addedge(int u,int v){    edge[edge_cnt].v=v;    edge[edge_cnt].next=head[u];    head[u]=edge_cnt++;}void tarjan(int u,int f){    dfn[u]=low[u]=++dfs_clock;    ins[u]=1; s[top++]=u;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==f)            continue;        if(!dfn[v])        {            fa[v]=u;            tarjan(v,u);            low[u]=min(low[u],low[v]);            if(low[v]>dfn[u])                bri[v]=1,bri_cnt++;        }        else if(ins[v])            low[u]=min(low[u],dfn[v]);    }    if(dfn[u]==low[u])    {        ins[u]=0;        while(s[--top]!=u)            ins[s[top]]=0;    }}void lca(int u,int v){    while(dfn[u]>dfn[v])    {        if(bri[u])            bri[u]=0,bri_cnt--;        u=fa[u];    }    while(dfn[v]>dfn[u])    {         if(bri[v])            bri[v]=0,bri_cnt--;        v=fa[v];    }    while(u!=v)    {        if(bri[u])            bri[u]=0,bri_cnt--;        if(bri[v])            bri[v]=0,bri_cnt--;        u=fa[u],v=fa[v];    }}int main(){    //freopen("text.txt","r",stdin);    int kase=0;    while(~scanf("%d%d",&n,&m) && n+m)    {        kase++;        printf("Case %d:\n",kase);        int u,v;        init();        for(int i=0;i<m;i++)        {            scanf("%d%d",&u,&v);            addedge(u,v); addedge(v,u);        }        tarjan(1,1);        scanf("%d",&k);        for(int i=0;i<k;i++)        {            scanf("%d%d",&u,&v);            lca(u,v);            printf("%d\n",bri_cnt);        }        printf("\n");    }    return 0;}