POJ 3694 Network(tarjan+lca)

Network Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 10463   Accepted: 3899 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 Source 2008 Asia Hefei Regional Contest Online by USTC

题意：

给你一个有向图，再依次给你q条边，依次添加，每次添加求出有几条桥。

POINT：

先把原始图求出桥，并且保存每个点的上一个点pre。如果a-b是桥，那么把isb[b]=1。

之后q个询问，找u和v的lca，途中经过的点如果有桥就--，并把isb=0。防止重复去桥。

注意他们的lca这个点是不能去桥的。

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxm=404000;
const int maxn=100100;
const int inf = 0x3f3f3f3f;
struct edge
{
    int v,used,nxt;
}len[maxm];
int DFN[maxn],low[maxn],pre[maxn];
int head[maxn],numl;
int isb[maxn];
int ans,tm;
void init()
{
    numl=0;
    ans=0;
    memset(head,-1,sizeof head);
    tm=0;
    memset(isb,0,sizeof isb);
    memset(DFN,0,sizeof DFN);
    memset(low,0,sizeof low);
}
void add(int u,int v)
{
    int &k=numl;
    len[k].v=v;
    len[k].nxt=head[u];
    len[k].used=0;
    head[u]=k++;
}
void lca(int u,int v)
{
    if(DFN[u]<DFN[v]) swap(u,v);
    while(DFN[u]>DFN[v])
    {
        if(isb[u]) isb[u]=0,ans--;
        u=pre[u];
    }
    while(u!=v)
    {
        if(isb[v]) isb[v]=0,ans--;
        v=pre[v];
    }
}
void dfs(int u)
{
    DFN[u]=low[u]=++tm;
    for(int i=head[u];i!=-1;i=len[i].nxt)
    {
        if(len[i].used||len[i^1].used) continue;
        len[i].used=len[i^1].used=1;
        int v=len[i].v;
        if(!DFN[v])
        {
            dfs(v);
            pre[v]=u;
            if(low[u]>low[v]) low[u]=low[v];
            if(low[v]>DFN[u])
            {
                isb[v]=1;
                ans++;
            }
        }
        else if(low[u]>DFN[v]) low[u]=DFN[v];
    }
}
int main()
{
    int n,m;
    int cas=0;
    while(~scanf("%d %d",&n,&m))
    {
        if(n==m&&m==0) break;
        init();
        int u,v;
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&u,&v);
            add(u,v);
            add(v,u);
        }
        int q;
        scanf("%d",&q);
        printf("Case %d:\n",++cas);
        pre[1]=0;
        dfs(1);
        for(int i=1;i<=q;i++)
        {
            scanf("%d %d",&u,&v);
            add(u,v);
            add(v,u);
            if(ans==0)
            {
                printf("0\n");
            }
            else
            {
                lca(u,v);
                printf("%d\n",ans);
            }
            
        }
        printf("\n");
        
    }
}