pku 3694 Network tarjan求割边

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

给你一个存在桥的无向连通图，每次增加一条边，问每次增加边之后还剩多少个桥。

朴素的办法就是每输入一次tarjan求一次桥的个数，肯定会tle；

正确的做发是，首先求出改图的桥的数量，在tarjan的过程中里用并查集把不是桥的边缩为一点，然后得到一棵树，每加入一条边，就肯定会形成一个环，，然后删除环中石桥的边，最后得出总的桥的数量。删除的过程就是检查是否石桥是就有cut--;

#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#define maxn 100010
using namespace std;
int low[maxn],dfn[maxn],father[maxn],f[maxn];
int index,n,m,q,cut;
vector<int>g[maxn];
void init(int nn)
{
    for (int i = 0; i <= n+1; ++i)
    {
        g[i].clear();
        low[i] = dfn[i] = father[i] = 0;
        f[i] = i;
    }
    index = cut = 0;
}
int find(int x)
{
    if (f[x] != x)
    f[x] = find(f[x]);
    return f[x];
}
int Union(int x,int y)
{
    x = find(x);
    y = find(y);
    if (x != y)
    {
        f[y] = x;
        return 1;
    }
    else
    return 0;

}
void tarjan(int i,int pre)
{
    int j,k;
    low[i] = dfn[i] = ++index;
    for (k = 0; k < g[i].size(); ++k)
    {
        j = g[i][k];
        if (!dfn[j])
        {
            tarjan(j,i);
            low[i] = min(low[i],low[j]);
            father[j] = i;//记录每个节点的父亲，方便后面的环中的查找
            if (low[j] > dfn[i])//记录桥数
             cut++;
            else
            Union(i,j);//把不是桥的缩为一点
        }
        else if (j != pre)//这里要注意不能是i的父亲，反之low[i] = low[father[i]]了;
        {
            low[i] = min(low[i],dfn[j]);
        }
    }
}
void lca(int u,int v)
{
    //在环中找桥的过程
    while (u != v)
    {
        while (dfn[u] >= dfn[v] && u != v)
        {
            if (Union(u,father[u]))
            cut--;
            u = father[u];
        }
        while (dfn[v] >= dfn[u] && u != v)
        {
            if (Union(v,father[v]))
            cut--;
            v = father[v];
        }
    }
}
int main()
{
     //freopen("d.txt","r",stdin);
    int i,x,y,cas = 1;
    while (scanf("%d%d",&n,&m))
    {
        if (!n && !m) break;
        printf("Case %d:\n",cas++);
        init(n);
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d",&x,&y);
            g[x].push_back(y);
            g[y].push_back(x);
        }
        /*for (i = 1; i <= n; ++i)
        {
            for (int j = 0; j < g[i].size(); ++j)
            printf("%d ",g[i][j]);
            puts(" ");
        }*/
        tarjan(1,-1);
        //printf(">>>%d\n",cut);
        cin>>q;
        while (q--)
        {
            cin>>x>>y;
            lca(x,y);
            printf("%d\n",cut);
        }
        printf("\n");
    }
    return 0;
}