洛谷1039 侦探推理

原题链接

讨厌模拟啊啊。。
因为不喜欢用\(string\)，所以用\(char\)数组打了一十分冗长的初始化。。

枚举今天是星期几及罪犯是谁，然后对每句话进行判断这个人讲的是真是假，如果没有冲突，那么最后计算说假话的总人数是否满足要求即可。
若有多种关于罪犯是谁的假设成立，那么就是不止一个罪犯；若没有假设成立，那么就是不可能有人为罪犯。

#include<cstdio>
#include<cstring>
using namespace std;
const int N = 110;
const int M = 25;
const int K = 260;
struct dd {
	int id, x, y;
};
dd tmy[N];
int nam_l[N], fake[M], nw_l, sen_l, n;
char sen[K], nam[M][N], nw[N];
char tr[6][M] = { "", "I am guilty.", "I am not guilty.", " is guilty.", " is not guilty.", "Today is " };
char da[8][M] = { "", "Monday.", "Tuesday.", "Wednesday.", "Thursday.", "Friday.", "Saturday.", "Sunday." };
bool e;
inline void re_nam(char *o, int &l)
{
	char c = getchar();
	for (; c < 'A' || c > 'Z'; c = getchar());
	for (l = 0; c != '\n' && c != '\r' && c != ' ' && c != ':'; c = getchar())
		o[l++] = c;
}
inline void re_sen(char o[], int &l)
{
	char c = getchar();
	for (; (c < 'A' || c > 'Z') && (c < 'a' || c > 'z'); c = getchar());
	for (l = 0; c != '\n' && c != '\r' && c > 0; c = getchar())
		o[l++] = c;
}
inline int fin()
{
	bool p;
	for (int i = 1; i <= n; i++)
		if (!(nam_l[i] ^ nw_l))
		{
			p = 0;
			for (int j = 0; j < nw_l && !p; j++)
				if (nam[i][j] ^ nw[j])
					p = 1;
			if (!p)
				return i;
		}
	return 0;
}
inline int fin_day()
{
	int i, j, l;
	for (i = 1; i <= 7; i++)
		if (!((l = strlen(da[i])) ^ nw_l))
		{
			for (j = 0; j < l; j++)
				if (da[i][j] ^ nw[j])
					break;
			if (j >= l)
				return i;
		}
	return 0;
}
inline bool chos(int i, int x, int y)
{
	int j, l = strlen(tr[i]);
	if ((y - x) ^ l)
		return false;
	for (j = 0; j < l; j++)
		if (sen[j + x] ^ tr[i][j])
			return false;
	return true;

}
inline void dm(int x, int y) { fake[x] >= 0 && fake[x] ^ y ? e = 1 : fake[x] = y; }
int main()
{
	int i, j, m, p, td, g = 0;
	scanf("%d%d%d", &n, &m, &p);
	for (i = 1; i <= n; i++)
		re_nam(nam[i], nam_l[i]);
	for (i = 1; i <= p; i++)
	{
		re_nam(nw, nw_l);
		tmy[i].id = fin();
		re_sen(sen, sen_l);
		if (chos(1, 0, sen_l))
			tmy[i].x = 1;
		else
			if (chos(2, 0, sen_l))
				tmy[i].x = 2;
			else
			{
				for (nw_l = j = 0; j < sen_l; j++)
					if (sen[j] < 'A' || sen[j] > 'Z')
						break;
					else
						nw[nw_l++] = sen[j];
				int k = fin();
				if (k && chos(3, j, sen_l))
					tmy[i].x = 3, tmy[i].y = k;
				else
					if (k && chos(4, j, sen_l))
						tmy[i].x = 4, tmy[i].y = k;
					else
						if (chos(5, 0, 9))
						{
							for (nw_l = 0, j = 9; j < sen_l; j++)
								nw[nw_l++] = sen[j];
							k = fin_day();
							if (k)
								tmy[i].x = 5, tmy[i].y = k;
						}
			}
	}
	for (i = 1; i <= n; i++)
		for (td = 1; td < 8; td++)
		{
			memset(fake, -1, sizeof(fake));
			e = 0;
			for (j = 1; j <= p && !e; j++)
			{
				int x = tmy[j].id, k = tmy[j].x;
				if (!k)
					continue;
				if (!(k ^ 1))
					dm(x, i ^ x ? 1 : 0);
				else
					if (!(k ^ 2))
						dm(x, i ^ x ? 0 : 1);
					else
						if (!(k ^ 3))
							dm(x, i ^ tmy[j].y ? 1 : 0);
						else
							if (!(k ^ 4))
								dm(x, i ^ tmy[j].y ? 0 : 1);
							else
								dm(x, td ^ tmy[j].y ? 1 : 0);
			}
			if (e)
				continue;
			int s_1 = 0, s_2 = 0;
			for (j = 1; j <= n; j++)
				if (fake[j] > 0)
					s_1++;
				else
					if (fake[j] < 0)
						s_2++;
			if (s_1 <= m && m <= s_1 + s_2)
			{
				if (g && g ^ i)
				{
					printf("Cannot Determine");
					return 0;
				}
				g = i;
			}
		}
	if (!g)
		printf("Impossible");
	else
		for (i = 0; i < nam_l[g]; i++)
			printf("%c", nam[g][i]);
	return 0;
}