算法知识点: 枚举,模拟,字符串处理
复杂度:
解题思路:
对于比较繁琐的模拟题,写代码的时候建议尽可能模块化。
依次枚举每个同学是否可能是凶手。最终结果有三种:
可能的凶手只有一个,输出凶手名字;
可能的凶手多于一个,输出"Cannot Determine";
任何一个同学都不可能是凶手,输出"Impossible";
在枚举每个同学是凶手的过程中,依次枚举今天是星期几,然后判断说谎的同学是否有可能有 n 个。
这里有两点需要注意:
每个同学需要始终如一,如果出现某个同学既说真话又说假话,那么当前枚举的情况不合法;
每个句子有三种情况:真话、假话、废话。最终只要 假话数量 ≤n≤ 假话数量 + 废话数量,那么说谎的同学就有可能有 n 个;
时间复杂度分析:
我们会依次枚举凶手、星期几以及句子,对于每个句子在判断是谁说的时会再循环一遍所有同学,所以总时间复杂度是,其中 M 是总人数,P 是总句子数。
C++ 代码:
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std; const int N = 110;
int n, m, P;
string sentence[N];
string name[N];
string weekday[7] = {
"Today is Monday.",
"Today is Tuesday.",
"Today is Wednesday.",
"Today is Thursday.",
"Today is Friday.",
"Today is Saturday.",
"Today is Sunday."
};
int state[N];
int get_person(string str)
{
for (int i = 0; i < m; i++)
if (name[i] == str)
return i;
return -1;
}
pair <int, string> get(string str)
{
int t = str.find(":");
int person = get_person(str.substr(0, t));
return make_pair(person, str.substr(t + 2));
}
int get_state(int bad_man, int day, int now, string line)
{
if (line == "I am guilty.")
{
if (bad_man == now) return 0;
return 1;
}
if (line == "I am not guilty.")
{
if (bad_man == now) return 1;
return 0;
}
int t = line.find(" is guilty.");
if (t != -1)
{
int p = get_person(line.substr(0, t));
if (p == bad_man) return 0;
return 1;
}
t = line.find(" is not guilty.");
if (t != -1)
{
int p = get_person(line.substr(0, t));
if (p != bad_man) return 0;
return 1;
}
for (int i = 0; i < 7; i++)
if (weekday[i] == line)
{
if (i == day) return 0;
return 1;
}
return -1;
}
bool check(int bad_man, int day)
{
memset(state, -1, sizeof state);
for (int i = 0; i < P; i++)
{
pair <int, string> t = get(sentence[i]);
int p = t.first;
int s = get_state(bad_man, day, p, t.second);
if (s == 0)
{
if (state[p] == 1) return false;
state[p] = s;
}
else if (s == 1)
{
if (state[p] == 0) return false;
state[p] = s;
}
}
int fake = 0, other = 0;
for (int i = 0; i < m; i++)
if (state[i] == 1)
fake++;
else if (state[i] == -1)
other++;
if (fake <= n && fake + other >= n) return true;
return false;
}
int main()
{
cin >> m >> n >> P;
for (int i = 0; i < m; i++) cin >> name[i];
getline(cin, sentence[0]);
for (int i = 0; i < P; i++) getline(cin, sentence[i]);
int cnt = 0, p;
for (int i = 0; i < m; i++)
{
bool flag = false;
for (int j = 0; j < 7; j++)
if (check(i, j))
{
flag = true;
break;
}
if (flag)
{
cnt++;
p = i;
}
}
if (cnt == 1) cout << name[p] << endl;
else if (cnt > 1) puts("Cannot Determine");
else puts("Impossible");
return 0;
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