1166 Summit – PAT甲级真题

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.

题目大意：为峰会安排休息区，一个理想的安排是邀请这些领导人，每个人互相之间都是直接朋友。给定一套暂定的安排，判断每个区域是否都已准备就绪。输入格式：第一行给一个正整数N，表示峰会的首领数量，以及一个正整数M，表示友谊关系的数量。接下来是M行，每行给出一对互为朋友的领导人的编号。领导人的编号从1到N。然后给出另一个正整数K，接下来是K行暂定的休息区，每行给出一个正整数L，然后是一系列L个不同的领导人编号。一行中所有数字都用空格分隔。输出格式：对于K个休息区的每一个，请按以下格式将您的建议输出在一行中：如果在这个休息区每个人都互相是直接朋友，并且没有朋友漏掉（即没有其他人是这个休息区每个人的直接朋友），就输出Area X is OK.如果在这个休息区每个人都是每个人的直接朋友，但在不破坏理想安排的情况下，可能还会邀请一些其他领导人，就输出Area X may invite more people, such as H. H是可以被邀请的领导人的最小编号。如果该休息区的安排不理想，则输出Area X needs help. 这样主持人可以提供一些特别的服务帮助领导人们互相了解。
分析：二维数组A作为邻接矩阵存储两个人是否是好朋友，如果是u和v好朋友就将数组A[u][v] = A[v][u] = 1；集合temp存储待检查的序列号。先检查所有的人是不是互相都为好朋友，如果不是的话，直接输出needs help。然后，检查剩下的所有人中，是否有人是他们所有人的好朋友、但是没有被邀请的，如果没有，就输出is OK. 否则输出may invite more people, such as f. 其中f为可以被邀请的领导人的最小编号~

#include <iostream>
#include <set>
using namespace std;
int n, m, k, l, u, v, s, e, f, f2, A[201][201];
int main() {
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        cin >> u >> v;
        A[u][v] = A[v][u] = 1;
    }
    cin >> k;
    for (int i = 1; i <= k; i++) {
        f = 0;
        set<int> temp;
        cin >> l;
        for (int j = 0; j < l; j++) {
            cin >> e;
            for (auto it : temp) 
                if (!A[e][it]) f = 1;
            temp.insert(e);
        }
        cout << "Area " << i;
        if (f) {
            cout << " needs help.\n";
        } else {
            for (int i = 1; i <= n; i++) {
                f2 = 1;
                if (temp.count(i)) continue;
                for (auto it : temp) {
                    if (!A[i][it]) {
                        f2 = 0;
                        break;
                    }
                }
                if (f2) {
                    f = i;
                    break;
                }
            }
            if (!f) 
                cout << " is OK.\n";
            else 
                cout << " may invite more people, such as " << f << ".\n"; 
        }
    }
    return 0;
}

#include <iostream>

#include <set>

using namespace std;

int n, m, k, l, u, v, s, e, f, f2, A[201][201];

int main() {

cin >> n >> m;

for (int i = 0; i < m; i++) {

cin >> u >> v;

A[u][v] = A[v][u] = 1;

}

cin >> k;

for (int i = 1; i <= k; i++) {

f = 0;

set<int> temp;

cin >> l;

for (int j = 0; j < l; j++) {

cin >> e;

for (auto it : temp)

if (!A[e][it]) f = 1;

temp.insert(e);

}

cout << "Area " << i;

if (f) {

cout << " needs help.\n";

} else {

for (int i = 1; i <= n; i++) {

f2 = 1;

if (temp.count(i)) continue;

for (auto it : temp) {

if (!A[i][it]) {

f2 = 0;

break;

}

if (f2) {

f = i;

break;

}

if (!f)

cout << " is OK.\n";

else

cout << " may invite more people, such as " << f << ".\n";

}

return 0;

}

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