题目链接:uva 11534 - Say Goodbye to Tic-Tac-Toe
题目大意:给定一个1*n的个子,每次操作可以选中一个未填过的个子画X或者O,如果该次操作形成了XX或者OO,那么该次操作者视为失败,人为先手,对于给定状态(注意当前状态也算在步数中),问是否可以战胜电脑。
解题思路:对于固定长度,两端的可能有空,X,O,组合情况共有9种,虽然有些情况等价,但是为方便处理,分为9种情况考虑。预先处理出各个长度下9种情况的SG值。然后对于给定状态,枚举位置放X和O,判断新状态的Nim和。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100;
int s[maxn+5][9];
int SG (int l, int x, int y) {
int vis[maxn+5];
memset(vis, 0, sizeof(vis));
for (int i = 1; i <= l; i++) {
if ((i != 1 || x != 1) && (i != l || y != 1)) {
int X = s[i-1][x*3+1] ^ s[l-i][1*3+y];
vis[X] = 1;
}
if ((i != 1 || x != 2) && (i != l || y != 2)) {
int O = s[i-1][x*3+2] ^ s[l-i][2*3+y];
vis[O] = 1;
}
}
int mv = -1;
while (vis[++mv]);
return mv;
}
void init (int n) {
memset(s[0], 0, sizeof(s[0]));
for (int i = 0; i < 9; i++)
s[1][i] = 1;
s[1][1*3+2] = s[1][2*3+1] = 0;
for (int i = 2; i <= n; i++) {
for (int x = 0; x < 3; x++) {
for (int y = 0; y < 3; y++)
s[i][x*3+y] = SG(i, x, y);
}
}
}
char str[maxn+5];
bool judge () {
int len = strlen(str+1);
int l = 0, r = 0, cnt = 0, ans = 0, k = 0;
for (int i = 1; i <= len; i++) {
if (str[i] == '.')
k++;
else {
r = (str[i] == 'X' ? 1 : 2);
ans ^= s[k][l*3+r];
cnt++;
l = r;
k = 0;
}
}
ans ^= s[k][l*3];
if (cnt&1)
ans = (ans == 0 ? 1 : 0);
return ans;
}
int main () {
init(maxn);
int cas;
scanf("%d", &cas);
while (cas--) {
scanf("%s", str+1);
printf("%s\n", judge() ? "Possible." : "Impossible.");
}
return 0;
}