1148 Werewolf - Simple Version (20 分)
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). Then Nlines follow and the i-th line gives the statement of the i-th player (1), 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 [ and [, if there exists 0 such that [ (i≤k) and [, 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
狼人杀，逻辑有点乱，一开始看错了题。题目大意：已知 N 名玩家中有 2 人扮演狼人角色，有 2 人说的不是实话，有狼人撒谎但并不是所有狼人都在撒谎。要求你找出扮演狼人角色的是哪几号玩家，如果有解，在一行中按递增顺序输出 2 个狼人的编号；如果解不唯一，则输出最小序列解；若无解则输出 No Solution～
1 #include <bits/stdc++.h>
2
3 using namespace std;
4 int n;
5 int s[105];
6 int ans[105];
7 int main(){
8 cin >> n;
9 for(int i = 1 ; i <= n; ++i)
10 cin >> s[i];
11 for(int i = 1; i <= n; i++){
12 for(int j = i+1; j <= n; j++){
13 for(int k = 1; k <= n; k++)
14 ans[k] = 1;
15 ans[i] = -1;ans[j] = -1;
16 vector<int> v;
17 for(int k = 1; k <= n; k++){
18 if(ans[abs(s[k])]*s[k] < 0){
19 v.push_back(k);
20 }
21 }
22 if(v.size() == 2&& ans[v[0]]+ans[v[1]] == 0){
23 cout <<i<< " " << j << endl;
24 return 0;
25 }
26 }
27 }
28 cout <<"No Solution"<<endl;
29 return 0;
30 }
原文地址：https://www.cnblogs.com/zllwxm123/p/11272785.html