挺有意思的一個建圖，夫妻之間男->女，前任之間女到男，然後判斷強連通分量

#include <bits/stdc++.h>
#define inf 2333333333333333
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(int i=a;i<=b;++i)
//by war
//2020.8.5
using namespace std;

int T;
int n,m,cnt,col,now,S,Max,tot;
int dfn[N],low[N],c[N],ans;
bool vis[N];
string
 
 x,y;
map<string,int>mp;
struct node{
    int n;
    node *next;
}*e[N];
stack<int>s;
struct edge{
    int x;int y;
}a[N];

void in(int &x){
    int y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1
 
)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(int x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

void push(int x,int y){
    node *p;
    p=new node();
    p->n=y;
    if(e[x]==0)
        e[x]=p;
    else{
        p->next=e[x]->next;
        e[x]->next=p;
    }
}

 
void tarjan(int x,int fa){
    dfn[x]=low[x]=++cnt;
    vis[x]=1;
    s.push(x);
    for(node *i=e[x];i;i=i->next){
        if(i->n==fa) continue;
        if(!dfn[i->n]){
            tarjan(i->n,x);
            low[x]=min(low[x],low[i->n]);
        }
        else
            if(vis[i->n]&&i->n!=fa)
                low[x]=min(low[x],dfn[i->n]);
    }
    if(low[x]==dfn[x]){
        col++;
        do{
            ans++;
            now=s.top();
            c[now]=col;
            s.pop();
            vis[now]=0;
        }while(x!=now);
    }
}

signed main(){
    in(n);
    For(i,1,n){
        cin>>x>>y;
        if(!mp[x]) mp[x]=++tot;
        if(!mp[y]) mp[y]=++tot;
        a[i].x=mp[x];
        a[i].y=mp[y];
        push(mp[x],mp[y]);
    }
    in(m);
    For(i,1,m){
        cin>>x>>y;
        push(mp[y],mp[x]);
    }
    For(i,1,tot) if(!dfn[i]) tarjan(i,i);
    For(i,1,n){
        if(c[a[i].x]==c[a[i].y]) puts("Unsafe");
        else puts("Safe");
    }
    return 0;
}