POJ-1904-King‘s Quest

链接：

https://vjudge.net/problem/POJ-1904

题意：

Once upon a time there lived a king and he had N sons. And there were N beautiful girls in the kingdom and the king knew about each of his sons which of those girls he did like. The sons of the king were young and light-headed, so it was possible for one son to like several girls.

So the king asked his wizard to find for each of his sons the girl he liked, so that he could marry her. And the king's wizard did it -- for each son the girl that he could marry was chosen, so that he liked this girl and, of course, each beautiful girl had to marry only one of the king's sons.

However, the king looked at the list and said: "I like the list you have made, but I am not completely satisfied. For each son I would like to know all the girls that he can marry. Of course, after he marries any of those girls, for each other son you must still be able to choose the girl he likes to marry."

The problem the king wanted the wizard to solve had become too hard for him. You must save wizard's head by solving this problem.

思路：

对王子喜欢的连一条边，再通过最后给的结婚关系，从女孩到王子连一条边。

再求强连通，再一个强连通内的王子和女孩肯定可以结婚。

代码：

#include <iostream>
#include <cstdio>
#include <vector>
#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
using namespace std;
const int MAXN = 4e3+10;

vector<int> G[MAXN];
int Dfn[MAXN], Low[MAXN];
int Vis[MAXN], Dis[MAXN];
int Fa[MAXN];
stack<int> St;
int n, cnt;
int times;

void Tarjan(int x)
{
    Dfn[x] = Low[x] = ++times;
    Vis[x] = 1;
    St.push(x);
    for (int i = 0;i < G[x].size();i++)
    {
        int nextnode = G[x][i];
        if (Dfn[nextnode] == 0)
        {
            Tarjan(nextnode);
            Low[x] = min(Low[x], Low[nextnode]);
        }
        else if (Vis[nextnode])
            Low[x] = min(Low[x], Dfn[nextnode]);
    }
    if (Low[x] == Dfn[x])
    {
        cnt++;
        while (St.top() != x)
        {
            Fa[St.top()] = cnt;
            Vis[St.top()] = 0;
            St.pop();
        }
        Fa[St.top()] = cnt;
        Vis[St.top()] = 0;
        St.pop();
    }
}

int main()
{
    scanf("%d", &n);
    for (int i = 1;i <= n;i++)
    {
        int num, g;
        scanf("%d", &num);
        while (num--)
        {
            scanf("%d", &g);
            G[i].push_back(n+g);
        }
    }
    int g;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d", &g);
        G[n+g].push_back(i);
    }
    for (int i = 1;i <= 2*n;i++)
    {
        if (Dfn[i] == 0)
            Tarjan(i);
    }
    for (int i = 1;i <= n;i++)
    {
        set<int> st;
        for (int j = 0;j < G[i].size();j++)
        {
            int node = G[i][j];
            if (Fa[i] == Fa[node])
                st.insert(node-n);
        }
        printf("%d", st.size());
        set<int>::iterator it = st.begin();
        while (it != st.end())
        {
            printf(" %d", *it);
            it++;
        }
        printf("\n");
    }

    return 0;
}