1022 Werewolf (35 分)（C++）

PAT顶级题解目录

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 vertion of this problem: given that there were N players, with M werewolves among them, at least one but not all the werewolves were lying, and there were exactly L 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 three positive integer N (5 ≤ N ≤ 100), M and L (2 ≤ M，L < N). 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 descending order the indices of the M 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 largest 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 larger than B. In case there is no solution, simply print No Solution.

Sample Input 1:

5 2 2
-2
+3
-4
+5
+4

Sample Output 1:

4 1

Sample Input 2:

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

Sample Output 2:

6 4

Sample Input 3:

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

Sample Output 3:

No Solution

题目大意：本题的简单版本在甲级中出现过，其他条件都类似，只是狼的数量和谎言家的数量不再是固定的。给出玩家数量n，狼的数量m，谎言家的数量L，现每个玩家说出一个数字，比如-2，代表该玩家认为2号是狼，若为2，则该玩家认为2号玩家是人。如果该号玩家说出的情况与实际不符合，则说明该玩家撒谎。现要找到一种满足狼的数量为m，谎言家的数量为L，且其中的狼不完全是谎言家（狼中谎言家的数量应该在[1,m)范围）。

解题思路：DFS剪枝，暴力遍搜狼，当遇到狼的数量为m的情况，则去查验该情况下谎言家的数量是否满足要求。

代码：

#include <bits/stdc++.h>
using namespace std;
int n, m, L, record[105];
bool iswolf[105] = {0}, flag = false; //分别存放该玩家是否是狼，以及是否已经找到符合条件的情况
std::vector<int> ans, tempans;
//判断谎言家是否满足要求
bool isTrue(){int cntL = 0, cntM_L = 0;for(int i = 1; i <= n; ++ i){if((iswolf[abs(record[i])] && record[i] > 0) || (!iswolf[abs(record[i])] && record[i] < 0)){++ cntL;if(iswolf[i])++ cntM_L; }}return cntL == L && cntM_L != m && cntM_L > 0;
}
void dfs(int index){if(flag)return;if(index == 0 && tempans.size() == m){if(isTrue()){flag = true;ans = tempans;}return;}if(tempans.size() > m || index == 0)return;for(int i = index; i > 0; -- i){tempans.push_back(i);iswolf[i] = 1;dfs(i-1);iswolf[i] = 0;tempans.pop_back();dfs(i-1);}
}
int main(){scanf("%d %d %d", &n, &m, &L);for(int i = 1; i <= n; ++ i)scanf("%d", &record[i]);dfs(n);if(!flag)printf("No Solution\n");else{printf("%d", ans[0]);for(int i = 1; i < ans.size(); ++ i)printf(" %d", ans[i]);}
}