pta 1148 Werewolf

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

题意：n个人玩狼人杀的游戏，每轮有两个狼人，有一个村民和一个狼人会说谎，求出两个狼人的编号。每个人会给出一个信息，正数代表村民，负数代表狼人。

开始想的有些复杂，后来发现直接暴力枚举就可以了，代码如下：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     int a[102];
 7     cin >> n;
 8     int wco = 0,pco = 0;
 9     for(int i = 1;i <= n;i++)
10     cin >> a[i];
11     int i,j,k;
12     for(i = 1;i <= n;i++)
13     {
14         for(j = i + 1;j <= n;j++)
15         {
16             bool flag = true;
17             wco = 0,pco = 0;
18             for(k = 1;k <= n;k++)
19             {
20                 if(a[k] < 0)//是狼
21                 {
22                     if(-a[k] != i && -a[k] != j)//其实不是狼，k说谎
23                     {
24                         if(k == i || k == j)// k是狼
25                         wco ++;
26                         if(k != i && k != j)//k是人
27                         pco ++;
28                         if(wco + pco > 2)
29                         {
30                             flag = false;
31                             break;
32                         }
33                     }
34                 }
35                 else//a[k]是人
36                 {
37                     if(a[k] == i || a[k] == j)//其实a[k]是狼，k说谎
38                     {
39                         if(k == i || k == j)// k是狼
40                         wco ++;
41                         if(k != i && k != j)//k是人
42                         pco ++;
43                         if(wco + pco > 2)
44                         {
45                             flag = false;
46                             break;
47                         }
48                     }
49                 }
50             }
51             if(flag && wco == 1 && pco == 1)
52             {
53                  cout << i << " " << j << endl;
54                  return 0;
55             }
56         }
57     }
58      cout << "No Solution" << endl;
59     return 0;
60 }