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;
}

UVA 10561 - Treblecross（博弈SG函数),布布扣,bubuko.com