本文主要是介绍POJ-1904 King's Quest 强连通分量求完美匹配，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
http://poj.org/problem?id=1904
题目意思：有n个女生和n个男生,给定一些关系表示男生喜欢女生(即两个人可以结婚),再给定一个初始匹配,表示这个男生和哪个女生结婚,初始匹配必定是合法的.求每个男生可以和哪几个女生可以结婚其他人还难能都找个女生结婚。
思路：第一反应还以为是求二分完美匹配，百度题解再知道用连通分量 0.0 将男生从1到n编号,女生从(n+1)到2*n编号,输入时如果男生u可以和女生v结婚,那么就做一条从u到v的边(u,v),对于输入的初始匹配(u,v)(表示男生u和女生v结婚),那么从v做一条到u的边(v,u),然后求解图的强连通分量,如果男身i和女生j在同一个强连通分量内,且i到j有直接的边。则他们可以结婚。
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn = 4105;
const int inf = 1<<29;
int n;
int times,ans,top;
bool vis[maxn];
int low[maxn],dfn[maxn],stack[maxn],p[maxn];
vector<int>map[maxn];
vector<int>cnt;void dfs(int u)
{int v;dfn[u] = low[u] = ++times; //为节点u设定次序vis[u] = true; //判断是否在栈中stack[++top] = u; //将节点i压入栈中for ( int i = 0; i < map[u].size(); i ++ ) //遍历每一条边{v = map[u][i];if ( !dfn[v] ) //如果节点i未被访问过{dfs(v); //继续向下查找low[u] = low[v] <= low[u]?low[v]:low[u];}else if ( vis[v] ) //如果节点i在栈中low[u] = dfn[v] <= low[u]?dfn[v]:low[u];}if ( dfn[u] == low[u] ) //如果节点i是强连通分量的根{ans++; //强连通分量数do{v = stack[top--]; //把栈里v上面的顶点弹栈p[v] = ans;vis[v] = false;}while ( v != u );}
}
void tarjan()
{ans = times = top = 0; //初始化memset( vis,0,sizeof(vis) );memset( dfn,0,sizeof(dfn) );for( int i = 1; i <= n; i ++ ) //遍历每个节点if ( !dfn[i] )dfs(i);
}
int main()
{//freopen("data.txt","r",stdin);int m,temp,g;while( scanf("%d",&m) != EOF ){for( int i = 1; i <= 2*m; i ++ )map[i].clear();for( int i = 1; i <= m; i ++ ){scanf("%d",&temp);for( int j = 1; j <= temp; j ++ ){scanf("%d",&g);map[i].push_back(g+m);}}for( int i = 1; i <= m; i ++ ){scanf("%d",&g);map[g+m].push_back(i);}n = m*2;tarjan();for( int u = 1; u <= m; u ++ ){cnt.clear();for( int i = 0; i < map[u].size(); i ++ ){int v = map[u][i];if( p[u] == p[v] )cnt.push_back(v-m);}printf("%d",cnt.size());sort( cnt.begin(),cnt.end() );for( int j = 0; j < cnt.size(); j ++ )printf(" %d",cnt[j]);puts("");}}return 0;
}#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
using namespace std;
const int maxn = 5105;
const int inf = 1<<29;
int n;
int times,ans,top;
bool map[maxn][maxn],vis[maxn];
int low[maxn],dfn[maxn],stack[maxn],p[maxn];
void dfs(int u)
{dfn[u] = low[u] = ++times; //为节点u设定次序vis[u] = true; //判断是否在栈中stack[++top] = u; //将节点i压入栈中for ( int i = 1; i <= n; i ++ ) //遍历每一条边{if( map[u][i] ){if (!dfn[i]) //如果节点i未被访问过{dfs(i); //继续向下查找low[u] = low[i] <= low[u]?low[i]:low[u];}else if ( vis[i] ) //如果节点i在栈中low[u] = dfn[i] <= low[u]?dfn[i]:low[u];}}int i;if ( dfn[u] == low[u] ) //如果节点i是强连通分量的根{ans++; //强连通分量数do{i = stack[top--]; //把栈里v上面的顶点弹栈p[i] = ans;vis[i] = false;}while ( i != u );}
}
void tarjan()
{int i;ans = times = top = 0; //初始化memset( vis,0,sizeof(vis) );memset( dfn,0,sizeof(dfn) );for( i = 1; i <= n; i ++ ) //遍历每个节点if ( !dfn[i] )dfs(i);
}
int main()
{//freopen("data.txt","r",stdin);int m,temp,g;while( scanf("%d",&m) != EOF ){memset(map,0,sizeof(map));for( int i = 1; i <= m; i ++ ){scanf("%d",&temp);for( int j = 1; j <= temp; j ++ ){scanf("%d",&g);map[i][m+g] = true;}}for( int i = 1; i <= m; i ++ ){scanf("%d",&g);map[m+g][i] = true;}n = m*2;tarjan();vector<int>ans;for( int i = 1; i <= m; i ++ ){ans.clear();for( int j = m+1; j <= n; j ++ )if( map[i][j] && p[i] == p[j] )ans.push_back(j-m);printf("%d",ans.size());for( int j = 0; j < ans.size(); j ++ )printf(" %d",ans[j]);puts("");}}return 0;
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