19年冬季第二题 PAT甲级 1166 Summit (25分)

7-3 Summit (25分)

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.where H is 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.

主要是写出流程图，做题之前1.仔细仔细仔细审题。2.建模。和那啥哈密顿图很像？还是啥图来着

//如果没有人和这里的人全部是朋友，且这里的人两两都是朋友，就是ok
//如果还有人和这里的所有人都是朋友，就直接输出这个人
//如果不全都是朋友，就输出need help
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std;
const int maxn = 205;
const int INF = 1000000000;
int G[maxn][maxn];int main(){fill(G[0],G[0]+maxn*maxn,INF);int n,m;//人是从0开始的 cin>>n>>m;int u,v;for(int i = 0;i < m; i++){cin>>u>>v;G[u][v] = G[v][u] = 1;}int q,k,dian;cin>>q;for(int i = 1; i <= q; i++){vector<int> v;cin>>k;bool needHelp = false;bool mayInvite = false;int inviteIndex;for(int j = 0; j < k; j++){cin>>dian;v.push_back(dian);}for(int j = 0; j < k -1;j++){for(int l = j+1;l < k; l++){if(G[v[j]][v[l]] == INF){needHelp = true;}}}for(int j = 1; j <= n; j++){bool flag = true;for(int l = 0; l < k; l++){if(G[j][v[l]] == INF){flag = false;break;}}if(flag == true){mayInvite = true;inviteIndex = j;break;}}if(needHelp == true) printf("Area %d needs help.\n",i);else if(mayInvite == true) printf("Area %d may invite more people, such as %d.\n",i,inviteIndex);else printf("Area %d is OK.\n",i);}return 0;
}

搞清楚判断的是G[v[j]][v[l]]不是j和l！

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