【PTA Advanced】1148 Werewolf - Simple Version（C++）

题目

Input Specification:

Output Specification:

Sample Input 1:

Sample Output 1:

Sample Input 2:

Sample Output 2 (the solution is not unique):

Sample Input 3:

Sample Output 3:

思路

代码

题目

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

思路

难度评级：⭐️⭐️⭐️⭐️

难度评四颗星的原因是，我拿到这题时一点思路也没有，不知道如何下手，看了PAT 1148 Werewolf 柳婼の blog的代码才知道其实是枚举每一种狼人情况的方法，并且柳神非常巧妙地运用了正负数乘积为负数这一点作为一些 if 的判断，非常值得学习借鉴。

代码

#include <iostream>
#include <vector>
#include <algorithm>using namespace std;int main(int argc, char** argv) {int n;cin>>n;vector<int> words(n+1);for(int i=1;i<=n;i++) cin>>words[i];// 枚举每一种狼人情况 for(int i=1;i<n;i++) {for(int j=i+1;j<=n;j++) {vector<int> identity(n+1, 1);// 标识每个人的身份，1表示村民，-1表示狼人 vector<int> liars; // 假设用户i和用户j为狼人identity[i]=-1;identity[j]=-1;// 遍历每个人说的话，判断他是否说谎 for(int k=1;k<=n;k++) {if(words[k]*identity[abs(words[k])]<0) {liars.push_back(k);}}// 检查说谎的人是否有两个人，且一个是狼人一个村民if(liars.size()==2&&identity[liars[0]]*identity[liars[1]]<0) {cout<<i<<" "<<j;return 0;} }}cout<<"No Solution";return 0;
}