ZOJ 3601 Unrequited Love【STL】

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Unrequited Love

Time Limit: 16 Seconds Memory Limit: 131072 KB

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

22 1 4BoyA 1 GirlCBoyB 1 GirlCGirlC 1 BoyA2 BoyA BoyB2 BoyA GirlC2 BoyB GirlC3 BoyA BoyB GirlC2 2 2H 2 O SHe 0O 1 HS 1 H3 H O S4 H He O S

Sample Output

001 BoyB000

注意：①king如果有只有一个，因为出现king就表示着剩余的人不是喜欢全部的人；②先假定第一个人a满足条件，从第二个人开始遍历。若不满足条件，则把a置成b，而对于跳过的人，必定存在一个他不爱的人，这就是当时的a，不然，a就要等于他。所以最后对a进行判断即可，包括爱与被爱的关系。

#include<cstdio>#include<math.h>#include<cstring>#include<climits>#include<string>#include<queue>#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<climits>#include<string>#include<queue>#include<stack>#include<set>#include<map>#include<algorithm>using namespace std;#define rep(i,j,k)for(i=j;i<k;i++)#define per(i,j,k)for(i=j;i>k;i--)#define MS(x,y)memset(x,y,sizeof(x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define ll long long#define abs(x) (x>0?x:-x)const int INF=0x7ffffff;const ll MAX=1e18;int main(){int t;cin>>t;while(t--){        set<string> x[30001];        map<string,int> id;        string z[30001];        string h,w;        int n,m,a,b;        int i,j,k;cin>>n>>m>>a;int p=-1;for(i=0;i<n+m;++i){cin>>h>>b;//if(id.count(h)==0)                id[h]=++p;for(j=0;j<b;++j){cin>>w;x[id[h]].insert(w);}}for(i=0;i<a;++i){cin>>b;for(j=0;j<b;++j)cin>>z[j];string g=z[0];int flag=0;for(j=1;j<b;++j){if(!x[id[g]].count(z[j])||x[id[z[j]]].count(g)){g=z[j];flag=j;}            }for(j=0;j<flag;++j){if(!x[id[g]].count(z[j])||x[id[z[j]]].count(g)){break;}}if(j!=flag)cout<<0<<endl;elsecout<<1<<" "<<g<<endl;}cout<<endl;    }return 0;}

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