pta--1148 Werewolf - Simple Version（20 分）（枚举逻辑）_综合

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 liers. 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 liers. 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

【分析】emmm这道题，，题意理解之后还要搞清楚怎么转化。感觉我短时间内想不到这种解决方法。感谢小伙伴给我讲清楚了~主要还是思维逻辑思维！

枚举i、j（表示两个狼人），假设的时候只要判断s[x]是否与i或j相等，即判断是不是狼人。lier和wolf分别表示说谎的人和狼人的数目，符合题意时只能lier=wolf=1；

【代码】

#include<bits/stdc++.h>
using namespace std;const int maxn=110;
int s[maxn];int main()
{int n;scanf("%d",&n);for(int i=1;i<=n;++i)scanf("%d",&s[i]);int lier,wolf;//说谎人数与狼人数for(int i=1;i<=n;++i){for(int j=i+1;j<=n;++j){//两个狼：i、jlier=wolf=0;int flag=1;for(int k=1;k<=n;++k){if(s[k]<0)//第k个人说这个人是狼 {if(-s[k]!=i && -s[k]!=j)//但这个人不是狼 {if(k!=i && k!=j)lier++;if(lier+wolf>2){flag=0;break;}if(k==i || k==j)wolf++;if(lier+wolf>2){flag=0;break;}}}else	//第k个人说这个人不是狼 {if(s[k]==i || s[k]==j)//但这个人是狼 {if(k!=i && k!=j)lier++;if(lier+wolf>2){flag=0;break;}if(k==i && k==j)wolf++;if(lier+wolf>2){flag=0;break;}}}}if(flag && lier==1 && wolf==1){printf("%d %d\n",i,j);return 0;}}} puts("No Solution");return 0;
}