A1166 Summit (25 分)

A1166 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:

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：

给来参加峰会的国家或政府首脑安排位置。

input:

各个政府首脑的关系图，编号是从 1~N
每个区域安排的位置

output：

判断每个区域安排的位置：
- 如果这个区域的人互相都是朋友，并且没有遗漏共同的朋友则输出Area X is OK.
- 如果这个区域的人互相都是朋友，并且遗漏了共同的朋友，则输出Area X may invite more people, such as H.，H表示遗漏的共同朋友中编号最小的那个。
- 如果这个区域的人只要出现有一对不是朋友关系，则输出Area X needs help.

思路

用hash table unordered_map<int, bool> 存储两个是否是朋友关系
遍历给出的位置安排，根据题目给出的条件判断

AC代码

#include <bits/stdc++.h>
using namespace std;

int n, m, k, t;
unordered_map<int, bool> heads;

int judge(vector<int> &arrange) {
    // -1: need help; 0: ok; other number invite people;
    int len = t;
    for (int i = 0; i < len - 1; ++i) {
        for (int j = i + 1; j < len; ++j) {
            if (!heads[arrange[i] * 1000 + arrange[j]]) {
                return -1;
            }
        }
    }
    unordered_set<int> st(arrange.begin(), arrange.end());
    // 寻找是否还有额外共同的朋友
    for (int i = 1; i <= n; ++i) {
        if (st.find(i) != st.end()) continue;
        bool b = true;
        for (int &h : arrange) {
            if (!heads[h * 1000 + i]) {
                b = false;
                break;
            }
        }
        if (b) return i;
    }
    return 0;
}

int main() {
    scanf("%d%d", &n, &m);
    int t1, t2;
    for (int i = 0; i < m; ++i) {
        cin >> t1 >> t2;
        heads[t1 * 1000 + t2] = true;
        heads[t2 * 1000 + t1] = true;
    }
    scanf("%d", &k);
    for (int i = 1; i <= k; ++i) {
        cin >> t;
        vector<int> arrange(t);
        for (int j = 0; j < t; ++j) {
            cin >> arrange[j];
        }
        int res = judge(arrange);
        if (res == 0) printf("Area %d is OK.\n", i);
        else if (res == -1) printf("Area %d needs help.\n", i);
        else printf("Area %d may invite more people, such as %d.\n", i, res);
    }
    return 0;
}