UVA 10561 - Treblecross（博弈SG函数)

来源：互联网发布：买家怎么进淘宝试衣间编辑：程序博客网时间：2024/05/29 03:45

UVA 10561 - Treblecross

题目链接

题意：给定一个串，上面有'X'和'.'，可以在'.'的位置放X，谁先放出3个'X'就赢了，求先手必胜的策略

思路：SG函数，每个串要是上面有一个X，周围的4个位置就是禁区了(放下去必败)，所以可以以X分为几个子游戏去求SG函数的异或和进行判断，至于求策略，就是枚举每个位置就可以了

代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;const int N = 205;int t, out[N], on, len, sg[N];char str[N];bool win() {for (int i = 0; i < len - 2; i++) {if (str[i] == 'X' && str[i + 1] == 'X' && str[i + 2] == 'X')return true; } return false;}int mex(int x) {bool vis[N];int i, t;if (sg[x] != -1) return sg[x];if (x == 0) return sg[x] = 0;memset(vis, false, sizeof(vis));for (int i = 1; i <= x; i++) {int t = mex(max(0, i - 3))^mex(max(0, x - i - 2));vis[t] = true; } for (int i = 0; i < N; i++) { if (vis[i]) continue;   return sg[x] = i;}}bool towin() {for (int i = 0; i < len; i++) {if (str[i] == '.') {str[i] = 'X';if (win()) {str[i] = '.';   return false;}str[i] = '.';  } } int ans = 0, num = 0; for (int i = 0; i < len; i++) { if (str[i] == 'X' || (i >= 1 && str[i - 1] == 'X') || (i >= 2 && str[i - 2] == 'X') || (i + 1 < len && str[i + 1] == 'X') || (i + 2 < len && str[i + 2] == 'X')) { ans ^= mex(num); num = 0;   }   else num++;  }  ans ^= mex(num);  return ans == 0;}void solve() {on = 0;len = strlen(str); for (int i = 0; i < len; i++) {  if (str[i] != '.') continue;  str[i] = 'X';  if (win() || towin())  out[on++] = i + 1;str[i] = '.';   }}int main() {memset(sg, -1, sizeof(sg));scanf("%d", &t);while (t--) {scanf("%s", str);solve();if (on == 0) printf("LOSING\n\n");else {printf("WINNING\n%d", out[0]);for (int i = 1; i < on; i++)printf(" %d", out[i]);printf("\n");  }}return 0;}

1 0