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

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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即可

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;}

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