(PAT 1148) Werewolf - Simple Version (枚举)

忘是亡心i 2022-03-17 13:58 123阅读 0赞

Werewolf（狼人杀） is a game in which the players are partitioned into two parties: the werewolves and the human beings. Suppose that in a game,

*  player \#1 said: "Player \#2 is a werewolf.";
 *  player \#2 said: "Player \#3 is a human.";
 *  player \#3 said: "Player \#4 is a werewolf.";
 *  player \#4 said: "Player \#5 is a human."; and
 *  player \#5 said: "Player \#4 is a human.".

Given that there were 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. Can you point out the werewolves?

Now you are asked to solve a harder version of this problem: given that there were N players, with 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. You are supposed to point out the werewolves.

### Input Specification: ###

Each input file contains one test case. For each case, the first line gives a positive integer N (5≤N≤100). Then N lines follow and the i-th line gives the statement of the i-th player (1≤i≤N), which is represented by the index of the player with a positive sign for a human and a negative sign for a werewolf.

### Output Specification: ###

If a solution exists, print in a line in ascending order the indices of the two werewolves. The numbers must be separated by exactly one space with no extra spaces at the beginning or the end of the line. If there are more than one solution, you must output the smallest solution sequence -- that is, for two sequences A=a\[1\],...,a\[M\] and B=b\[1\],...,b\[M\], if there exists 0≤k<M such that a\[i\]=b\[i\] (i≤k) and a\[k+1\]<b\[k+1\], then A is said to be smaller than B. In case there is no solution, simply print `No Solution`.

### Sample Input 1: ###

5
    -2
    +3
    -4
    +5
    +4

### Sample Output 1: ###

1 4

### Sample Input 2: ###

6
    +6
    +3
    +1
    -5
    -2
    +4

### Sample Output 2 (the solution is not unique): ###

1 5

### Sample Input 3: ###

5
    -2
    -3
    -4
    -5
    -1

### Sample Output 3: ###

No Solution

解题思路:

这题想了一下午没有思路,参考了人家的思路后恍然大悟

[https://www.liuchuo.net/archives/6494][https_www.liuchuo.net_archives_6494]

首先我们定义一个数组words用来存储玩家说的话

然后开始枚举狼人,每次枚举两人,设人类为1,狼人为-1,存到一个数组identity中

然后遍历验证 如果 word\[k\]\*identity\[abs(words\[k\])\] < 0; 那么这个人就是在说谎,把说谎的人存到一个数组中

如果说谎的人有两个,且1个为人类另一个为狼人,那么就输出结果

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <string>
    #include <vector>
    using namespace std;
    const int MAXN = 110;
    int N;
    int thEidentity[MAXN];
    int words[MAXN];
    int groupCount = 0;
    int main() {
    	int num;
    	scanf("%d", &N);
    	for (int i = 1; i <= N; ++i) {   //全部从1开始计算
    		scanf("%d", &num);
    		words[i] = num;
    	}
    	//开始暴力假设
    	for (int i = 1; i <= N; ++i) {
    		for (int j = i+1; j <= N; ++j) {
    			fill(thEidentity, thEidentity + MAXN, 1);
    			thEidentity[i] = thEidentity[j] = -1;   //枚举狼人
    			vector<int> liervec;
    			for (int k = 1; k <= N; ++k) {
    				if ((words[k] * thEidentity[abs(words[k])]) < 0) {
    					liervec.push_back(k);
    				}
    			}
    			if (liervec.size() == 2 && thEidentity[liervec[0]] + thEidentity[liervec[1]] == 0) {
    				cout << i << " " << j << endl;
    				system("PAUSE");
    				return 0;
    			}
    		}
    	}
    	cout << "No Solution" << endl;
    	system("PAUSE");
    	return 0;
    }

[https_www.liuchuo.net_archives_6494]: https://www.liuchuo.net/archives/6494