A summit (峰會) 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.?題目大意：為峰會安排休息區，一個理想的安排是邀請這些領導人，每個人互相之間都是直接朋友。給定一套暫定的安排，判斷每個區域是否都已準備就緒。抽象一下可以看做有m條邊，n個頂點的無向圖，給定k次查詢，每次查詢給出點集中點的數量，以及頂點編號，問這些點是否兩兩連通，如果不聯通輸出needs help；如果是的話，是否存在其它頂點加入后仍然保持兩兩連通，有這樣的點的話輸出最小的點編號，否則輸出OK。
分析：由于點的數量很少，可以直接用暴力的方法檢查是否是連通子圖，如果是的話再檢查是否有其它點加入后仍然連通。
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <cmath>
#include <map>
#include <set>
#define INF 0xffffffff
#define db1(x) cout<<#x<<"="<<(x)<<endl
#define db2(x,y) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<endl
#define db3(x,y,z) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<endl
#define db4(x,y,z,r) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<endl
#define db5(x,y,z,r,w) cout<<#x<<"="<<(x)<<", "<<#y<<"="<<(y)<<", "<<#z<<"="<<(z)<<", "<<#r<<"="<<(r)<<", "<<#w<<"="<<(w)<<endl
using namespace std;int main(void)
{#ifdef testfreopen("in.txt","r",stdin);//freopen("in.txt","w",stdout);clock_t start=clock();#endif //testint n,m;scanf("%d%d",&n,&m);int num[n+5][n+5];for(int i=0;i<=n;++i)for(int j=0;j<=n;++j)num[i][j]=0;for(int i=0;i<m;++i){int a,b;scanf("%d%d",&a,&b);num[a][b]=num[b][a]=1;}int k;scanf("%d",&k);for(int Case=1;Case<=k;++Case){printf("Area %d ",Case);int cnt,f=1;scanf("%d",&cnt);int ques[cnt+5]={0};for(int i=0;i<cnt;++i){scanf("%d",&ques[i]);if(i&&f){for(int j=0;j<i;++j){if(!num[ques[i]][ques[j]]){f=0;break;}}}}if(!f)printf("needs help.\n");else{int index=1;for(int i=1;i<=n;++i){f=1;index=i;for(int j=0;j<cnt;++j){if(num[i][ques[j]]==0){f=0;break;}}if(f)break;}if(!f)printf("is OK.\n");else printf("may invite more people, such as %d.\n",index);}}#ifdef testclockid_t end=clock();double endtime=(double)(end-start)/CLOCKS_PER_SEC;printf("\n\n\n\n\n");cout<<"Total time:"<<endtime<<"s"<<endl; //s為單位cout<<"Total time:"<<endtime*1000<<"ms"<<endl; //ms為單位#endif //testreturn 0;
}
)
