POJ3694(求割边)

Network

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7943		Accepted: 2893

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

题意:给定结点和边确定一幅无向图，然后增加Q条边，输出每增加一条边之后图中的割边数目。
思路：先利用tarjan求割边。新增加的一条边两端的结点u、v，u和v到它们的LCA之间的割边全部消失。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=100005;
struct Edge{
    int to,net;
}es[MAXN*4];
int head[MAXN],tot;
void addedge(int u,int v)
{
    es[tot].to=v;
    es[tot].net=head[u];
    head[u]=tot++;
}

int n,m,q;
int dfn[MAXN],low[MAXN],key;
bool bridge[MAXN];
int par[MAXN],depth[MAXN];
int cnt;
void tarjan(int u,int fa,int dep)
{
    par[u]=fa;
    depth[u]=dep;
    dfn[u]=++key;
    low[u]=key;
    for(int i=head[u];i!=-1;i=es[i].net)
    {
        int to=es[i].to;
        if(!dfn[to])
        {
            tarjan(to,u,dep+1);
            low[u]=min(low[u],low[to]);
            if(dfn[u]<low[to])
            {
                bridge[to]=true;
                cnt++;
            }
        }
        else if(to!=fa)    low[u]=min(low[u],dfn[to]);
    }
}

void query(int u,int v)
{
    if(depth[u]>depth[v])    swap(u,v);
    while(depth[v]>depth[u])
    {
        if(bridge[v])
        {
            bridge[v]=false;
            cnt--;
        }
        v=par[v];
    }
    while(u!=v)
    {
        if(bridge[u])
        {
            bridge[u]=false;
            cnt--;
        }
        u=par[u];
        
        if(bridge[v])
        {
            bridge[v]=false;
            cnt--;
        }
        v=par[v];
    }
}

int main()
{
    int cas=0;
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)!=0)
    {
        cnt=0;
        memset(head,-1,sizeof(head));
        key=0;
        memset(dfn,0,sizeof(dfn));
        memset(low,0,sizeof(low));
        memset(bridge,false,sizeof(bridge));
        for(int i=0;i<m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        tarjan(1,0,0);
        scanf("%d",&q);
        printf("Case %d:\n",++cas);
        while(q--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            query(u,v);
            printf("%d\n",cnt);
        }
        printf("\n");
    }
    return 0;
}