UVA 11534 - Say Goodbye to Tic-Tac-Toe(博弈sg函数)

UVA 11534 - Say Goodbye to Tic-Tac-Toe

题目链接

题意：给定一个序列，轮流放XO，要求不能有连续的XX或OO。最后一个放的人赢。问谁赢

思路：sg函数。每一段...看成一个子游戏，利用记忆化求sg值，记忆化的状态要记录下左边和右边是X还是O就可以

代码：

#include <stdio.h>
#include <string.h>

const int N = 105;
int t, sg[3][3][N];
char str[N];

int getnum(char c) {
	if (c == 'X') return 1;
 	if (c == 'O') return 2;
}

int mex(int s, int e, int l) {
	if (sg[s][e][l] != -1) return sg[s][e][l];
	if (l == 0) return sg[s][e][l] = 0;
	bool vis[N];
	memset(vis, false, sizeof(vis));
	for (int i = 1; i <= l; i++) {
		for (int j = 1; j <= 2; j++) {
			if (i == 1 && s == j) continue;
			if (i == l && e == j) continue;
			int t = mex(s, j, i - 1)^mex(j, e, l - i);
			vis[t] = true;
  		}
 	}
 	for (int i = 0; ;i++)
 		if (!vis[i]) return sg[s][e][l] = i;
}

int main() {
	memset(sg, -1, sizeof(sg));
	scanf("%d", &t);
	while (t--) {
		scanf("%s", str);

		int len = strlen(str), s = 0, e = 0, l = 0, ans = 0, cnt = 0;
		for (int i = 0; i < len; i++) {
			if (str[i] == '.')
				l++;
			else {
				e = getnum(str[i]);
				ans ^= mex(s, e, l);
				s = e; l = 0; cnt++;
   			}
  		}
  		ans ^= mex(s, 0, l);
  		if (cnt&1)
    		ans = (ans == 0?1:0);
  		printf("%s\n", ans?

"Possible.":"Impossible.");
	}
	return 0;
}