POJ 1904 King's Quest 强联通分量+输入输出外挂

题意：国王有n个儿子，现在这n个儿子要在n个女孩里选择自己喜欢的，有的儿子可能喜欢多个，最后国王的向导给出他一个匹配。匹配有n个数，代表某个儿子和哪个女孩可以结婚。已知这些条件，要你找出每个儿子可以和哪些女孩结婚

思路：求强联通分量。同时练习一下输入输出外挂可以减少时间

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<string>
#include<cmath>
#include<vector>
using namespace std;
const int maxn=1e5+7;
const double eps=1e-8;
const double pi=acos(-1);
const int inf = 0x3f3f3f3f;
#define ll long long
#define clc(a,b) memset(a,b,sizeof(a))

const int N=4005;
const int M=204000;
struct Edge {
    int v,next;
} edge[M];

int first[N],low[N],dfn[N],sta[M],belong[N],ans[N];
bool instack[N];
int g,cnt,top,scc;

void addedge(int u,int v) {
    edge[g].v=v;
    edge[g].next=first[u];
    first[u]=g++;
}

int min(int a,int b) {
    return a<b?a:b;
}

void tarjan(int u) {
    int i,v;
    low[u]=dfn[u]=++cnt;
    sta[++top]=u;
    instack[u]=true;
    for(i=first[u]; i!=-1; i=edge[i].next) {
        v=edge[i].v;
        if(!dfn[v]) {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        } else if(instack[v]) {
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]) {
        scc++;
        while(1) {
            v=sta[top--];
            instack[v]=false;
            belong[v]=scc;
            if(u==v)
                break;
        }
    }
}

int scan() {
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}

void out(int a) {
    if(a>9)
        out(a/10);
    putchar(a%10+'0');
}

int main() {
    int n,i,u,v,k;
    while(scanf("%d",&n)!=EOF) {
        g=cnt=top=scc=0;
        clc(first,-1);
        clc(dfn,0);
        clc(instack,false);
        for(i=1; i<=n; i++) {
            k=scan();
            while(k--) {
                v=scan();
                addedge(i,v+n);
            }
        }
        for(i=1;i<=n;i++){
            v=scan();
            addedge(v+n,i);
        }
         for(i=1;i<=2*n;i++){
            if(!dfn[i])
                tarjan(i);
         }
         for(u=1;u<=n;u++){
            int countt=0;
            for(i=first[u];i!=-1;i=edge[i].next){
                v=edge[i].v;
                if(belong[u]==belong[v])
                    ans[countt++]=v-n;
            }
            sort(ans,ans+countt);
            out(countt);
            for(i=0;i<countt;i++){
                putchar(' ');
                out(ans[i]);
            }
            putchar('
');
         }
    }
    return 0;
}