Network（Tarjan+缩点+LCA）

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

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;
}