1148 Werewolf - Simple Version （20 分) 暴力模拟

àì夳堔傛蜴生んèń 2022-02-26 09:35 117阅读 0赞

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≤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

看了很久的题才读懂狼人杀规则，一定有2个人在撒谎，且一个为狼人一个为村民，n个人中有且仅有2个为狼人。

n<100，所以可以枚举2个狼人和那个说谎的村民。

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<sstream>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    int main()
    {
    	int n,i,j,k,a[105],f=0;
    	scanf("%d",&n);
    	for(i=1;i<=n;i++)
    	scanf("%d",&a[i]);
    	for(i=1;i<=n;i++)
    	{
    		for(j=i+1;j<=n;j++)
    		{
    			vector<int> lie;//存说谎的人 
    			int v[105];
    			for(k=1;k<=n;k++)v[k]=1;
    			v[i]=v[j]=-1;//设i和j是狼人 
    			for(k=1;k<=n;k++)
    			{
    				if(a[k]*v[abs(a[k])]<0)//与事实不符，则判为说谎
    				lie.push_back(k);
    			}
    			if(lie.size()==2&&v[lie[0]]+v[lie[1]]==0)//说谎的人一个为狼人，一个为村民
    			{
    				printf("%d %d\n",i,j);
    				f=1;
    				break;
    			}
    		}
    		if(f)break;
    	}
    	if(f==0)
    	printf("No Solution\n");
    }