Find the Winning Move
| Time Limit: 3000MS | Memory Limit: 32768K | |
| Total Submissions: 1286 | Accepted: 626 |
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
#include<cstdio>
using namespace std;
char s[5][5];
int chess,X,Y;
inline int abs(int x){return x>0?x:-x;}
bool check(int x,int y){//判断一个局面是否结束
int tot=0;
for(int i=0;i<4;i++) s[x][i]==‘o‘?tot++:s[x][i]==‘x‘?tot--:tot;//横向判断
if(abs(tot)==4) return 1;tot=0;
for(int i=0;i<4;i++) s[i][y]==‘o‘?tot++:s[i][y]==‘x‘?tot--:tot;//纵向判断
if(abs(tot)==4) return 1;tot=0;
for(int i=0;i<4;i++) s[i][i]==‘o‘?tot++:s[i][i]==‘x‘?tot--:tot;//正对角线判断
if(abs(tot)==4) return 1;tot=0;
for(int i=0;i<4;i++) s[i][3-i]==‘o‘?tot++:s[i][3-i]==‘x‘?tot--:tot;//反对角线判断
if(abs(tot)==4) return 1;
return 0;
}
int Min(int ,int);
int Max(int ,int);
int Max(int x,int y){
if(check(x,y)) return -1;//已经结束(对方胜)
if(chess==16) return 0;//平局
for(int i=0,now;i<4;i++){
for(int j=0;j<4;j++){
if(s[i][j]==‘.‘){
s[i][j]=‘x‘;chess++;
now=Min(i,j);
s[i][j]=‘.‘;chess--;
//对方需要找的最差估价，如果当前比之前最差的高，α剪枝
if(now==1) return 1;
}
}
}
return -1;
}
int Min(int x,int y){
if(check(x,y)) return 1;//已经结束(己方胜)
if(chess==16) return 0;
for(int i=0,now;i<4;i++){
for(int j=0;j<4;j++){
if(s[i][j]==‘.‘){
s[i][j]=‘o‘;chess++;
now=Max(i,j);
s[i][j]=‘.‘;chess--;
//自己需要找的最高估价，如果当前比之前最差的低，β剪枝
if(!now||now==-1) return -1;
}
}
}
return 1;
}
bool solve(){
for(int i=0,now;i<4;i++){
for(int j=0;j<4;j++){//枚举，然后搜索
if(s[i][j]==‘.‘){
s[i][j]=‘x‘;chess++;
now=Min(i,j);
s[i][j]=‘.‘;chess--;
if(now==1){
X=i;Y=j;
return 1;
}
}
}
}
return 0;
}
int main(){
char ch[3];
while(~scanf("%s",ch)&&ch[0]==‘?‘){
for(int i=0;i<4;i++) scanf("%s",s[i]);chess=0;
for(int i=0;i<4;i++) for(int j=0;j<4;j++) chess+=s[i][j]!=‘.‘;
if(chess<=4){puts("#####");continue;}//一定平局(对方都绝顶聪明的话)
if(solve()) printf("(%d,%d)\n",X,Y);
else puts("#####");
}
return 0;
}