A summit (峰会) is a meeting of heads of state or government. Arranging the rest areas for the summit is not a simple job. The ideal arrangement of one area is to invite those heads so that everyone is a direct friend of everyone.
Now given a set of tentative arrangements, your job is to tell the organizers whether or not each area is all set.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (≤ 200), the number of heads in the summit, and M, the number of friendship relations. Then M lines follow, each gives a pair of indices of the heads who are friends to each other. The heads are indexed from 1 to N.
Then there is another positive integer K (≤ 100), and K lines of tentative arrangement of rest areas follow, each first gives a positive number L (≤ N), then followed by a sequence of L distinct indices of the heads. All the numbers in a line are separated by a space.
Output Specification:
For each of the K areas, print in a line your advice in the following format:
-
if in this area everyone is a direct friend of everyone, and no friend is missing (that is, no one else is a direct friend of everyone in this area), print
Area X is OK.. -
if in this area everyone is a direct friend of everyone, yet there are some other heads who may also be invited without breaking the ideal arrangement, print
Area X may invite more people, such as H.whereHis the smallest index of the head who may be invited. -
if in this area the arrangement is not an ideal one, then print
Area X needs help.so the host can provide some special service to help the heads get to know each other.
Here X is the index of an area, starting from 1 to K.
Sample Input:
8 10
5 6
7 8
6 4
3 6
4 5
2 3
8 2
2 7
5 3
3 4
6
4 5 4 3 6
3 2 8 7
2 2 3
1 1
2 4 6
3 3 2 1
Sample Output:
Area 1 is OK.
Area 2 is OK.
Area 3 is OK.
Area 4 is OK.
Area 5 may invite more people, such as 3.
Area 6 needs help.
Meaning of the questions:
To N points, M edges. The K inquiry. Each inquiry is given L points, ask the L twenty-two point is not connected.
If any two connected:
A memory other do not exist, are connected to the L points:
Have:Area i may invite more people, such as 这个点.
No:Area i is OK.
Twenty-two not connected:Area i needs help.
answer:
AC Code:
#include<bits/stdc++.h> using namespace std; int e[205][205]; int a[205]; int v[205]; int n,m; int main(){ cin>>n>>m; memset(e,0,sizeof(e)); for(int i=1;i<=m;i++){ int u,v; cin>>u>>v; e[u][v]=e[v][u]=1;//Storing adjacency matrix } int K; CIN >> K; int NUM; for ( int I = . 1 ; I <= K; I ++) { // K th interrogation CIN >> NUM; Memset (V, 0 , the sizeof (V) ); // V asked to mark num points for ( int J = . 1 ; J <= num; J ++ ) { CIN >> a [J]; V [a [J]] = . 1 ; // cook flag } int F = . 1; // is not twenty-two connected for ( int J = . 1 ; J <= NUM; J ++ ) { for ( int P = J + . 1 ; P <= NUM; P ++ ) { IF (E [A [J]] [A ! [P]] = . 1 ) F = 0 ; BREAK ; } } IF (F) COUT <<! " Area " << I << " Needs Help. " ; the else { // If pairwise connected int ANS = - 1 ; // if there is for ( int J = . 1 ; J <= n-; J ++) { // query exists or not a point which points are connected num IF (V [J]) Continue ; // se num point is not in the int FF = . 1 ; for ( int P = . 1 ; P <= NUM; P ++ ) { IF (E [A [P]] [J] =! . 1 ) { FF = 0 ; BREAK ; } } IF (FF) { // meet with each point are connected in num ans=j;//存在 break; } } if(ans!=-1){//存在 cout<<"Area "<<i<<" may invite more people, such as "<<ans<<"."; }else{//不存在 cout<<"Area "<<i<<" is OK."; } } if(i!=k) cout<<endl;// end of the no blank lines } return 0 ; }