[题解] [SDOI2010] 大陆争霸
作者:互联网
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\n", bl[girl[i]] == bl[boy[i]] ? "Unsafe" : "Safe");
return 0;
}
标签:争霸,now,int,题解,SDOI2010,read,num,mp,include 来源: https://www.cnblogs.com/ztlztl/p/11184465.html