POJ 3694 Network （求桥，边双连通分支缩点，lca）

Network

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5619		Accepted: 1939

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

 

 

 

 

 

 

这题是给了一个连通图。

 

问再加入边的过程中，桥的个数。

 

 

先对原图进行双连通分支缩点。可以形成一颗树。

 

 

这颗树的边都是桥。

然后加入边以后，查询LCA，LCA上的桥都减掉。

 

标记边为桥不方便，直接标记桥的终点就可以了。

 

 

具体看代码吧！
很好的题目

 

 

 

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;

const int MAXN = 100010;
const int MAXM = 400010;

struct Edge
{
    int to,next;
    bool cut;
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];//Belong数组的值是1~block
int Index,top;
int block;
bool Instack[MAXN];
int bridge;

void addedge(int u,int v)
{
    edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;
    head[u] = tot++;
}
void Tarjan(int u,int pre)
{
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u];i != -1;i = edge[i].next)
    {
        v = edge[i].to;
        if( v == pre )continue;
        if( !DFN[v] )
        {
            Tarjan(v,u);
            if(Low[u] > Low[v])Low[u] = Low[v];
            if(Low[v] > Low[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
        }
        else if(Instack[v] && Low[u] > DFN[v])
             Low[u] = DFN[v];
    }
    if(Low[u] == DFN[u])
    {
        block++;
        do
        {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = block;
        }
        while( v != u );
    }
}
void init()
{
    tot = 0;
    memset(head,-1,sizeof(head));
}

vector<int>vec[MAXN];
int father[MAXN];
int dep[MAXN];
int a[MAXN];
void lca_bfs(int root)
{
    memset(dep,-1,sizeof(dep));
    dep[root] = 0;
    a[root] = 0;//桥的标记，标记桥的一个点
    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);
        }
    }
}
int ans;
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;
        }
        u = father[u];
        v = father[v];
    }
}
void solve(int N)
{
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    Index = top = block = 0;
    Tarjan(1,1);
    for(int i = 1;i <= block;i++)
      vec[i].clear();
    for(int u = 1;u <= N;u++)
        for(int i = head[u];i != -1;i = edge[i].next)
           if(edge[i].cut)
           {
               int v = edge[i].to;
               vec[Belong[u]].push_back(Belong[v]);
               vec[Belong[v]].push_back(Belong[u]);
           }
    lca_bfs(1);
    ans = block - 1;
    int Q;
    int u,v;
    scanf("%d",&Q);
    while(Q--)
    {
        scanf("%d%d",&u,&v);
        lca(Belong[u],Belong[v]);
        printf("%d
",ans);
    }
    printf("
");
}
int main()
{
    int n,m;
    int u,v;
    int iCase = 0;
    while(scanf("%d%d",&n,&m)==2)
    {
        iCase++;
        if(n==0 && m == 0)break;
        init();
        while(m--)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        printf("Case %d:
",iCase);
        solve(n);
    }
    return 0;
}