PAT(甲级)1148 Werewolf - Simple Version(枚举)

Description

狼人杀游戏，第几个人说第几个人是狼人或者是人，规定所有人中有两个狼人，其中只有一个狼人和一个人说谎，要求找出两个狼人

Input

第一行n，代表n个人
接下来n行
正数 D_i 代表：第i个人说第D_i人是human
负数 D_i 代表：第i个人说第D_i（绝对值）人是werewolf

Output

输出两个werewolf的下标，一个空格，如果有多个答案，输出最小的序列

解题思路

算法标签：枚举
显然就是枚举任意两个人是狼人，统计n个人说谎的人数，如果说话人数为两人，且一个为狼人另一个为人，当然没有规定这两个人谁是狼人谁是人，只是要求其中一个是狼人，另一个是人就行，遍历从小到大遍历，第一个找到的自然是最小的，无任何算法，不需要任何技巧，简单遍历即可(没玩过狼人杀但是可以百度啊！没吃过猪肉还没见过猪跑嘛)

代码

//freopen("hao.txt","r",stdin);
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
const int N = 105;
int main()
{int n = 0;int statement[N] = {0};bool isans = false;// ture为human，false为werewolfbool iswolf[N];cin>>n;for(int i=1;i<=n;i++)cin>>statement[i];for(int i=1;i<n;i++) {for(int j=i+1;j<=n;j++) {// 初始化均为humanmemset(iswolf,true,sizeof(iswolf));// 创建说谎人的队列,存储下标queue<int>liar;//cout<<i<<" "<<j<<endl;// 认为第i个人和第j个人是狼人，这样就可以保证答案从小到大iswolf[i] = false;iswolf[j] = false;// 寻找有几个人是说谎的，以及说谎人是不是一个werewolf和humanfor(int k=1;k<=n;k++) {if((statement[k]>0)&&(iswolf[abs(statement[k])]==false))liar.push(k);else if((statement[k]<0)&&(iswolf[abs(statement[k])]==true))liar.push(k);}if(liar.size()==2) {int liar_a = liar.front();int liar_b = liar.back();if(iswolf[liar_a]==true && iswolf[liar_b]==false) {cout<<i<<" "<<j;isans = true;break;}else if(iswolf[liar_a]==false && iswolf[liar_b]==true) {cout<<i<<" "<<j;isans = true;break;}}}if(isans)break;}// 判断是否有答案if(!isans)cout<<"No Solution";return 0;
}

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