PKU1568 Find the Winning Move

Find the Winning Move
Time Limit:3000MS? Memory Limit:32768K
Total Submit:120 Accepted:67

Description
4x4 tic-tac-toe is played on a board with four rows (numbered 0 to 3 from top to bottom) and four columns (numbered 0 to 3 from left to right). There are two players, x and o, who move alternately with x always going first. The game is won by the first player to get four of his or her pieces on the same row, column, or diagonal. If the board is full and neither player has won then the game is a draw.
Assuming that it is x's turn to move, x is said to have a forced win if x can make a move such that no matter what moves o makes for the rest of the game, x can win. This does not necessarily mean that x will win on the very next move, although that is a possibility. It means that x has a winning strategy that will guarantee an eventual victory regardless of what o does.

Your job is to write a program that, given a partially-completed game with x to move next, will determine whether x has a forced win. You can assume that each player has made at least two moves, that the game has not already been won by either player, and that the board is not full.

Input
The input contains one or more test cases, followed by a line beginning with a dollar sign that signals the end of the file. Each test case begins with a line containing a question mark and is followed by four lines representing the board; formatting is exactly as shown in the example. The characters used in a board description are the period (representing an empty space), lowercase x, and lowercase o. For each test case, output a line containing the (row, column) position of the first forced win for x, or '#####' if there is no forced win. Format the output exactly as shown in the example.

Output
For this problem, the first forced win is determined by board position, not the number of moves required for victory. Search for a forced win by examining positions (0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), ..., (3, 2), (3, 3), in that order, and output the first forced win you find. In the second test case below, note that x could win immediately by playing at (0, 3) or (2, 0), but playing at (0, 1) will still ensure victory (although it unnecessarily delays it), and position (0, 1) comes first.

Sample Input

?
....
.xo.
.ox.
....
?
o...
.ox.
.xxx
xooo
$

Sample Output

#####
(0,1)

Source
Mid-Central USA 1999

这个博弈 我用了Alpha-Beta剪枝 搜索是标准的极大极小的搜索过程
hash用的是4进制的便于位运算的hash
Solution
//by oyjpArt
#include <stdio.h>
#include <string.h>
char brd[4][4];
const int MAX = (1<<24)+1;
char hash[MAX];
char init[4][4];

bool win(char a) {
?//hor
?int i, j;
?for(i = 0; i<4; i++) {
??for(j = 0; j<4; j++)
???if(brd[i][j] != a)
????break;
??if(j == 4) return 1;
?}
?//ver
?for(i = 0; i<4; i++) {
??for(j = 0; j<4; j++)
???if(brd[j][i] != a)
????break;
??if(j == 4) return 1;
?}
?//dia
?if(brd[0][0] == a&& brd[1][1] == a && brd[2][2] == a && brd[3][3] == a) return 1;
?if(brd[0][3] == a&& brd[1][2] == a && brd[2][1] == a && brd[3][0] == a) return 1;
?return 0;
}

int res() {
?int i, j;
?if(win('x')) return 2;
?if(win('o')) return -2;
?for(i = 0; i<4; i++)
?? for(j = 0; j<4; j++)
??? if(brd[i][j] == '.')
???? return 1;
?return 0;
}

int hash_brd() { //4进制 便于2进制的hash
?int i, j, idx = 0, addr = 0;
?for(i = 0; i<4; i++)
??for(j = 0; j<4; j++)
???if(init[i][j] == '.') {
????if(brd[i][j] == 'o') ?addr |= 1<<(idx*2);
????else if(brd[i][j] == 'x')?addr |= 1<<(idx*2+1);
????idx++;
???}
?return addr;
}

int search(int player, int sta) {
?int hashbrd = hash_brd();
?if(hash[hashbrd] != -1) return hash[hashbrd];
?int i, j, tmp;
?int result = res();
?if(result != 1) {
??hash[hashbrd] = result;
??return result;
?}
?int min = 100, max = -100;
?if(player == 0) { //对方选min 如果子局面搜到了一个比当前min还大的 就不用搜了
??for(i = 0; i<4; i++) {
???for(j = 0; j<4; j++)
????if(brd[i][j] == '.') {
?????brd[i][j] = 'o';
?????if( (tmp = search(1, min)) < min ) min = tmp;
?????brd[i][j] = '.';
?????if(min == -2 || min <= sta) break;
????}
???if(j != 4) break;
??}
??hash[hashbrd] = min;
??return min;
?}
?else {
??for(i = 0; i<4; i++) {
???for(j = 0; j<4; j++)
????if(brd[i][j] == '.') {
?????brd[i][j] = 'x'; //如果子局面搜索时搜到比当前max小的 就不用搜了
?????if( (tmp = search(0, max)) > max) max = tmp;
?????brd[i][j] = '.';
?????if(max >= sta || max == 2) break;
????}
???if(j != 4) break;
??}
??hash[hashbrd] = max;
??return max;
?}
}

????
int main() {
//?freopen("t.in", "r", stdin);
?char tmp[2];
?int i, j;
?while(scanf("%s", tmp), strcmp(tmp, "$") != 0) {
??memset(hash, -1, sizeof(hash));
??for(i = 0; i<4; i++)
???scanf("%s", brd[i]);
??memcpy(init, brd, sizeof(brd));
??for(i = 0; i<4; i++) {
???for(j = 0; j<4; j++)
????if(brd[i][j] == '.') {
?????brd[i][j] = 'x';
?????if(search(0, -100) == 2) {
??????printf("(%d,%d)\n", i, j);
??????break;
?????}
?????brd[i][j] = '.';
????}
???if(j != 4) break;
??}
??if(i == 4) printf("#####\n");
?}
?return 0;
}

?