A1166 Summit（25分）PAT甲级 (Advanced Level) Practice（C++）满分题解【图】

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:

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.

代码如下：

#include<iostream>
#include<vector>
using namespace std;vector<vector<int>> edge(250, vector<int>(250, 0));int main() {int n, m;cin >> n >> m;//n个节点m条边for (int i = 0; i < m; i++) {int x, y;cin >> x >> y;edge[x][y] = edge[y][x] = 1;//建立图的连接}int k;cin >> k;for (int i = 1; i <= k; i++) {//查询的记录个数int l;//每个area的点的个数cin >> l;vector<int> v(l);for (int j = 0; j < l; j++)cin >> v[j];bool flag1 = true;for (int j = 0; j < l; j++) {for (int k = j + 1; k < l; k++) {//cout << v[j] << " " << v[k] << " " << edge[v[j]][v[k]] << endl;if (edge[v[j]][v[k]] == 0) {//两点无边，即没关系flag1 = false;break;}}if (flag1 == false) break;//已经不是理想安排了，直接break}bool flag2 = false;int minn = -1;for (int j = 1; j <= n; j++) {//遍历所有节点int k;for (k = 0; k < l; k++) {if (j == v[k]) break;//该点已经在area中，直接breakif (j != v[k] && edge[j][v[k]] == 0) break;//不在area中，但是与area中的当前点没有边，所以不符合，直接break}if (k == l) {//没有通过break退出，说明所有节点遍历完了，既符合不在area中，又与area中已有的所有节点有边，可以插入安排中flag2 = true;minn = j;//由于节点是从序号从小到大遍历的，而插入节点要输出序号最小的，所以找到的即符合要求break;}}if (flag1 == false) printf("Area %d needs help.\n", i);else if (flag2) printf("Area %d may invite more people, such as %d.\n", i, minn);else printf("Area %d is OK.\n", i);}
}

运行结果如下：