poj 3352 Road Construction(点双连通分量缩点+缩点树变为双连通分量)

来源：互联网发布：吾易抢票软件编辑：程序博客网时间：2024/06/05 15:30

题目链接：

点击打开链接

题目大意：

给出一张图，问最少加多少条边，将他变成边双连通图

题目分析：

首先进行点双连通图缩点，（点双连通图一定是边双连通图），然后得到一棵树，对于一棵树，我们只需要知道树的最底层有多少个点，然后将他们两两连接即可，那么所有的点就至少有两条路径到达，因为树的两条链相连就变成了环，环路就是路径上的点都有两条路径到达，然后很轻易的就能得到结果，缩点什么的早就是模板了

代码如下:

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <vector>#include <map>#include <stack>#define MAX 1007using namespace std;int n,m,u,v;int dfn[MAX],low[MAX],b[MAX],times,cnt;int in[MAX];typedef pair<int,int> pii;stack<int> s;vector<int> e[MAX];map<pii,bool> mp;void tarjan ( int u , int p ){    dfn[u] = low[u] = ++times;    s.push ( u );    int len = e[u].size();    for ( int i = 0 ; i < len ; i++ )    {        int v = e[u][i];        if ( v == p ) continue;        if ( !dfn[v] )        {            tarjan ( v , u );            low[u] = min ( low[u] , low[v] );        }        else            low[u] = min ( low[u] , dfn[v] );    }    if ( low[u] == dfn[u] )    {        int temp;        do        {            temp = s.top();            b[temp] = cnt;            s.pop();        }while ( temp != u );        cnt++;    }}void init ( ){    for ( int i = 0 ; i < MAX ; i++ )        e[i].clear();    while ( !s.empty())        s.pop();    times = cnt = 0;    mp.clear();    memset ( in , 0 , sizeof ( in ));}int main ( ){    while ( ~scanf ( "%d%d" , &n , &m ) )    {        init();        while ( m-- )        {            scanf ( "%d%d" , &u , &v );            e[u].push_back ( v );            e[v].push_back ( u );        }        for ( int i = 1 ; i <= n ; i++ )            if ( !dfn[i] ) tarjan ( i ,-1 );        for ( int i = 1 ; i <= n ; i++ )        {            int len = e[i].size();            for ( int j = 0 ; j < len ; j++ )            {                int v = e[i][j];                if ( b[i] == b[v] ) continue;                if ( mp[make_pair(i,v)] ) continue;                mp[make_pair(i,v)] = true;                in[b[v]]++;            }        }        int ans = 0;        for ( int i = 0 ; i < cnt ; i++ )            if ( in[i] == 1 ) ans++;        printf ( "%d\n" , (ans+1)/2 );    }}

0 0