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+;
 const double eps=1e-;
 const double pi=acos(-);
 const int inf = 0x3f3f3f3f;
 #define ll long long
 #define clc(a,b) memset(a,b,sizeof(a))

 const int N=;
 const int M=;
 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!=-; 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() {
             v=sta[top--];
             instack[v]=false;
             belong[v]=scc;
             if(u==v)
                 break;
         }
     }
 }

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

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

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