[题解] [国家集训队] 稳定婚姻

题面

题解

考虑在什么情况下此夫妻离婚后仍有n对夫妻

将男性看做黑点, 女性看做白点, 情侣关系和夫妻关系看做边

则若这对夫妻在一个黑白交错, 情侣关系和夫妻关系交错的一个环上(画图理解一下)

这对夫妻就是不安全的

考虑将边定向, 婚姻关系为女向男连边, 情侣关系为男向女连边

则若夫妻都在同一个强连通分量中这对夫妻关系就是不安全的

tarjan求强连通分量即可

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#define N 10005
using namespace std;

int cnt, tot, num, n, m, fa[N], girl[N], boy[N], head[N], bl[N], dfn[N], low[N], stk[N], top; 
struct edge { int to, next; } e[N << 2]; 
map<string, int> mp;
bool is[N]; 

inline int read()
{
	int x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w; 
}

inline void adde(int u, int v) { e[++tot] = (edge) { v, head[u] }; head[u] = tot; }

int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }

void tarjan(int u, int fa)
{
	dfn[u] = low[u] = ++tot; stk[++top] = u; is[u] = 1;
	for(int i = head[u]; i; i = e[i].next)
	{
		int v = e[i].to;
		if(!dfn[v]) tarjan(v, u), low[u] = min(low[u], low[v]);
		else if(v != fa && is[v]) low[u] = min(low[u], dfn[v]); 
	}
	if(low[u] >= dfn[u])
	{
		num++; 
		while(1)
		{
			int now = stk[top--];
			bl[now] = num; is[now] = 0; 
			if(now == u) break; 
		}
	}
}

int main()
{
	n = read();
	for(int i = 1; i <= n << 1; i++) fa[i] = i;
	for(int i = 1; i <= n; i++)
	{
		string G, B; cin>>G>>B; 
		if(!mp[G]) mp[G] = girl[i] = ++cnt;
		if(!mp[B]) mp[B] = boy[i] = ++cnt; 
		adde(boy[i], girl[i]); 
	}
	m = read();
	for(int i = 1; i <= m; i++)
	{
		string G, B; cin>>G>>B;
		adde(mp[G], mp[B]); 
	}
	tot = 0; 
	for(int i = 1; i <= cnt; i++)
		if(!dfn[i]) tarjan(i, 0);
	for(int i = 1; i <= n; i++)
		printf("%s
", bl[girl[i]] == bl[boy[i]] ? "Unsafe" : "Safe"); 
	return 0; 
}