TOJ 4289 Unrequited Love

Description

There are n single boys and m single girls. Each of them may love none, one or several of other people unrequitedly and one-sidedly. For the coming q days, each night some of them will come together to hold a single party. In the party, if someone loves all the others, but is not loved by anyone, then he/she is called king/queen of unrequited love.

Input

There are multiple test cases. The first line of the input is an integer T ≈ 50 indicating the number of test cases.

Each test case starts with three positive integers no more than 30000 -- n m q. Then each of the next n lines describes a boy, and each of the next m lines describes a girl. Each line consists of the name, the number of unrequitedly loved people, and the list of these people's names. Each of the last q lines describes a single party. It consists of the number of people who attend this party and their names. All people have different names whose lengths are no more than 20. But there are no restrictions that all of them are heterosexuals.

Output

For each query, print the number of kings/queens of unrequited love, followed by their names in lexicographical order, separated by a space. Print an empty line after each test case. See sample for more details.

Sample Input

2
2 1 4
BoyA 1 GirlC
BoyB 1 GirlC
GirlC 1 BoyA
2 BoyA BoyB
2 BoyA GirlC
2 BoyB GirlC
3 BoyA BoyB GirlC
2 2 2
H 2 O S
He 0
O 1 H
S 1 H
3 H O S
4 H He O S

Sample Output

0
0
1 BoyB
0

0
0

Source

ZJCPC 2012

不会写，看似水题的东西没想到这么难。

假如一个人存在unrequited love，那么表示这个结点没有入度，且这个点到所有的点一定是有路径的。

来自大牛的解题思路，数据比较的大用vector数组来建图，并且要使用二分找来判断是否存在通路。

#include <stdio.h>
#include <map>
#include <string>
#include <vector>
#include <algorithm>
#include <iostream>
#define MAXN 60005
using namespace std;

vector< int > V[MAXN];
vector< int > VP;
map< string , int > M;
string Name[MAXN];
int cnt;

void initVP(){
    VP.clear();
}

void initGraph(){
    for(int i=0; i<MAXN; i++){
        V[i].clear();
    }
}

void addString(string s){
    if(M.find(s)==M.end()){
        M[s]=cnt;
        Name[cnt++]=s;     
    }
}

void createMap(int n){
    char ch[21];
    string s;
    int c,u,v;
    for(int i=0; i<n; i++){
        scanf("%s" ,ch);
        s=string(ch);    
        addString(s);
        u=M[s];
        scanf("%d" ,&c);
        for(int j=0; j<c; j++){
            scanf("%s" ,ch);
            s=string(ch);
            addString(s);
            v=M[s];    
            V[u].push_back(v);
        }
        sort(V[u].begin(),V[u].end());
    }    
}

int main()
{
    int c,t;
    int n,m,q;
    scanf("%d" ,&t);
    while( t-- ){
        M.clear();
        cnt=0;
        initGraph();
        scanf("%d %d %d" ,&n ,&m ,&q);
        createMap(n);
        createMap(m);
        while( q-- ){
            initVP();
            char ch[21];
            string s;
            scanf("%d" ,&c);
            while( c-- ){
                scanf("%s" ,ch);
                s=string(ch);
                VP.push_back(M[s]);
            }
            int size=VP.size();
            int i,j;
            for(i=0; i<size; i++){
                for(j=0; j<size; j++){
                    if(i==j)continue;
                     if( !binary_search( V[VP[i]].begin() , V[VP[i]].end(), VP[j])  
                     || binary_search( V[VP[j]].begin(), V[VP[j]].end(), VP[i]))
                         break; 
                }
                if(j==size){
                    printf("1 %s
",Name[VP[i]].c_str());
                    break;
                }
            }
            if(i==size){
                puts("0");
            }
        }
        puts("");    
    }
    return 0;
}