UVA 10561 - Treblecross（博弈SG函数)

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
, on, len, sg
;
char str
;

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
;
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;
}