C++趣味算法之侦探推理

#include using namespace std;

//m：总人数 n：始终说谎人数 p：说话的总数int m, n, p;//judge[i]：第i句话是真是假，真1假-1不清楚0 w[i]：第i局话是编号多少的的人说的int judge[21], w[200];//err：矛盾标记 nx：当前可能的罪犯int err, nx;//name[i]：所有人名字（编号为1~m） say[i]：所有人说的话 day[i]：所有星期几名字string name[100], say[200];string day[10] = {"0", "Today is Sunday.", "Today is Monday.","Today is Tuesday.", "Today is Wednesday.", "Today is Thursday.","Today is Friday.", "Today is Saturday.",};

//sset函数标记一个人说话真假void sset(int who, int x) {if (judge[who] == -x) err = 1; //如果一个人既说真话又说假话，则矛盾else judge[who] = x;}

int main() {cin >> m >> n >> p;for (int i = 1; i <= m; i++) {cin >> name[i];}for (int i = 1; i <= p; i++) {string nm;cin >> nm; //输入说这句话人的名字nm.erase(nm.end() - 1); //删除nm中冒号，便于判断这句话编号多少的的人说的for (int j = 1; j <= m; j++) {if (name[j] == nm) w[i] = j;}getline(cin, say[i]);say[i].erase(say[i].begin()); //删除say[i]中的起始空格}

for (int td = 1; td <= 7; td++) { //暴力枚举今天是星期几for (int px = 1; px <= m; px++) { //暴力枚举罪犯编号是几号err = 0; //清除标记memset(judge, 0, sizeof(judge)); //初始化为不清楚真假//依次判断每一句说的话for (int i = 1; i <= p; i++) {int who = w[i]; //说这句话人的编号//如果一个人是罪犯，并且说自己是罪犯，则说的就是真话，否则就是假话if (say[i] == "I am guilty.") sset(who, px == who ? 1 : -1);//如果一个人不是罪犯，并且说自己不是罪犯，则说的就是真话，否则就是假话if (say[i] == "I am not guilty.") sset(who, px != who ? 1 : -1);//如果一个人说今天是星期几，说对了就是真话，说错了就是假话for (int j = 1; j <= 7; j++) {if (say[i] == day[j]) sset(who, j == td ? 1 : -1);}//如果一个人说其他人不是罪犯，说对了就是真话，说错了就是假话for (int j = 1; j <= m; j++) {if (say[i] == name[j] + " is guilty.") sset(who, j == px ? 1 : -1);if (say[i] == name[j] + " is not guilty.") sset(who, j != px ? 1 : -1);}}int cnt = 0; //说假话的人数int no = 0; //不清楚真假的的人数for (int i = 1; i <= m; i++) {if (judge[i] == -1) //假cnt++;if (judge[i] == 0) //不清楚no++;}//如果出现Impossible的情况，err = 1，出现矛盾//如果cnt<=n<=cnt+no,即假设合理if (!err && cnt <= n && cnt + no >= n) {if (nx && nx != px) { //如果出现了两个合理的罪犯cout << "Cannot Determine";return 0;} else {nx = px;}}}}if (!nx) cout << "Impossible";else cout << name[nx];