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.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:
8 105 67 86 43 64 52 38 22 75 33 464 5 4 3 63 2 8 72 2 31 12 4 63 3 2 1Sample 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.题目限度：
题目粗心：
现给定一个N个顶点，M条边的无向图，给出K个查问，每一个查问是一个顶点汇合，须要判断以后顶点联合是否是一个齐全子图，如果不是，输入Area X needs help. 其中X为查问的编号，从1开始，否则判断是否存在一个顶点与该汇合中的所有顶点都边相连，如果有，输入Area X may invite more people, such as H. 其中H为那个顶点。如果没有，输入Area X is OK。
算法思路：
应用邻接矩阵G判断任意两点是否连通，arrange存储每次查问的顶点汇合，isExist标记查问的顶点汇合，每一次查问的时候，首先应用isAllConnected判断arrange中的每一个点是否都有边相连，如果是返回true，否则返回false，printf("Area %d needs help.\n",i);,代码如下：
bool isAllConnected(){ for(int i:arrange){ for(int j:arrange){ if(i!=j&&G[i][j]==0){ return false; } } } return true;}而后再应用getMorePeople判断以后的顶点汇合是否能够再增加其他人退出，如果能够返回顶点编号，否则返回-1，如果返回-1，printf("Area %d is OK.\n",i);否则printf("Area %d may invite more people, such as %d.\n",i,a); (a为返回值),代码如下：
int getMorePeople(){ for (int i = 1; i <= N; ++i) { // 判断以后人i是否能够增加到汇合arrange中 bool possible = true; // i在汇合arrange中了 if(isExist[i]) continue; for(int j:arrange){ if(G[i][j]==0){ possible = false; } } if(possible){ // i退出汇合arrange中后与所有人都连通 return i; } } return -1;}提交后果：
AC代码：
#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<vector>using namespace std;int G[205][205];vector<int> arrange;int N,M;bool isExist[205];bool isAllConnected(){ for(int i:arrange){ for(int j:arrange){ if(i!=j&&G[i][j]==0){ return false; } } } return true;}int getMorePeople(){ for (int i = 1; i <= N; ++i) { // 判断以后人i是否能够增加到汇合arrange中 bool possible = true; // i在汇合arrange中了 if(isExist[i]) continue; for(int j:arrange){ if(G[i][j]==0){ possible = false; } } if(possible){ // i退出汇合arrange中后与所有人都连通 return i; } } return -1;}int main(){ scanf("%d %d",&N,&M); for (int i = 0; i < M; ++i) { int a,b; scanf("%d %d",&a,&b); G[a][b] = G[b][a] = 1; } int K; scanf("%d",&K); for(int i=1;i<=K;++i){ int num; scanf("%d",&num); arrange.clear(); memset(isExist,0, sizeof(isExist)); for (int j = 0; j < num; ++j) { int a; scanf("%d",&a); arrange.push_back(a); isExist[a] = true; } // 判断是否齐全连通 if(!isAllConnected()){ // 不是齐全连通 printf("Area %d needs help.\n",i); } else { int a = getMorePeople(); if(a==-1){ printf("Area %d is OK.\n",i); } else { printf("Area %d may invite more people, such as %d.\n",i,a); } } } return 0;}