Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Consider the problem called Fun Coloring below.

Fun Coloring Problem

Instance: A finite set U and sets S₁, S₂, S₃,…,S_m  U and |S_i| ≤ 3.

Problem: Is there a function f : U {RED, BLUE} such that at least one member of each S_i is assigned a different color from the other members?

Given an instance of Fun Coloring Problem, your job is to find out whether such function f exists for the given instance.

Input

In this problem U = {x₁, x₂, x₃,…,x_n}. There are k instances of the problem. The first line of the input file contains a single integer k and the following lines describe k instances, each instance separated by a blank line. In each instance the first line contains two integers n and m with a blank in between. The second line contains some integers i’s representing x_i’s in S₁, each i separated by a blank. The third line contains some integers i’s representing x_i’s in S₂ and so on. The line m+2 contains some integers i’s representing x_i’s in S_m. Following a blank line, the second instance of the problem is described in the same manner and so on until the k^th instance is described. In all test cases, 1 ≤ k ≤ 13, 4 ≤ n ≤ 22, and 6 ≤ m ≤ 111.

Output

For each instance of the problem, if f exists, print a Y. Otherwise, print N. Your solution will contain one line of k Y’s (or N’s) without a blank in between. The first Y (or N) is the solution for instance 1. The second Y (or N) is the solution for instance 2, and so on. The last Y (or N) is the solution for instance k.

 

 

Sample input	Sample output
2 5 3 1 2 3 2 3 4 1 3 5   7 7 1 2 1 3 4 2 4 3 2 3 1 4 5 6 7	YN

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窮舉所有 RED 與 GREEN 的放置方法，接著去查看每個集合不能全為 RED 或者 全為 GREEN。


#include <iostream>
#include <cstdio>
#include <sstream>
using namespace std;

int main() {
    int t, n, m, i, j, k;
    int sub[120][30], subt[120];
    cin >> t;
    string line;
    while(t--) {
        cin >> n >> m;
        getline(cin, line);
        for(i = 0; i < m; i++) {
            getline(cin, line);
            stringstream sin;
            sin << line;
            subt[i] = 0;
            while(sin >> sub[i][subt[i]])
                sub[i][subt[i]]--, subt[i]++;
        }
        int flag = 0;
        for(i = (1<<n)-1; i >= 0; i--) {
            for(j = 0; j < m; j++) {
                int cnt = 0;
                for(k = 0; k < subt[j]; k++)
                    cnt += (i>>sub[j][k])&1;
                if(cnt == 0 || cnt == subt[j])
                    break;
            }
            if(j == m) {
                flag = 1;
                break;
            }
        }
        if(flag)    putchar('Y');
        else        putchar('N');
    }
    return 0;
}