CodeForces - 825B Five-In-a-Row

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本文介绍了一种五子棋游戏的算法实现，通过遍历空位并检查放置棋子后是否能形成五个连续的棋子来判断游戏胜负。文章详细展示了如何通过水平、垂直及两个对角线方向检查棋盘状态。

Alice and Bob play 5-in-a-row game. They have a playing field of size 10 × 10. In turns they put either crosses or noughts, one at a time. Alice puts crosses and Bob puts noughts.

In current match they have made some turns and now it's Alice's turn. She wonders if she can put cross in such empty cell that she wins immediately.

Alice wins if some crosses in the field form line of length not smaller than 5. This line can be horizontal, vertical and diagonal.

Input

You are given matrix 10 × 10 (10 lines of 10 characters each) with capital Latin letters 'X' being a cross, letters 'O' being a nought and '.' being an empty cell. The number of 'X' cells is equal to the number of 'O' cells and there is at least one of each type. There is at least one empty cell.

It is guaranteed that in the current arrangement nobody has still won.

Output

Print 'YES' if it's possible for Alice to win in one turn by putting cross in some empty cell. Otherwise print 'NO'.

Examples

input
XX.XX.....
.....OOOO.
..........
..........
..........
..........
..........
..........
..........
..........

output
YES

input
XXOXX.....
OO.O......
..........
..........
..........
..........
..........
..........
..........
..........

output
NO

题意：加一个‘ X ’棋子，判断是否有五个‘ X ’能够连成一条线，只能放在‘ . ’上不能放在‘ 0 ’上，可以连成一条线输出‘YES’，不能输出‘NO’；

思路：把‘ . ’一个个枚举出来，然后判断它的上方，下方；左方，右方；左上方，右下方；右上方，左下方；能否五个

连成一线。

代码：

#include<stdio.h>
char map[15][15];
int main()
{
    int i,j,k,q;
    for(i=0; i<10; i++)
        scanf("%s",map[i]);
    for(i=0; i<10; i++)
    {
    	for(j=0; j<10; j++)
        {
            if(map[i][j]=='.') 
            {
                int s=1;//初始化为1； 
                for(int k=j+1; k<10; k++)//右方向 
                {
                    if(map[i][k]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                for(int k=j-1; k>=0; k--) //左方向 
                {
                    if(map[i][k]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                
                s=1;            
                for(int k=i+1; k<10; k++)   //上方 
                {
                    if(map[k][j]=='X')
                    {
                        s++;
                        if(s==5)
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                for(int k=i-1; k>=0; k--)  //下方 
                {
                    if(map[k][j]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                
                s=1;           
                for(k=i-1,q=j-1; q>=0,k>=0; q--,k--) //左上方 
                {
                    if(map[k][q]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                for(k=i+1,q=j+1; q<10,k<10; q++,k++)//右下方 
                {
                    if(map[k][q]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                
                s=1;        
                for(k=i+1,q=j-1; q>=0,k<10; q--,k++)// 右上方 
                {
                    if(map[k][q]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
                for(k=i-1,q=j+1; q<10,k>=0; q++,k--)//左下方 
                {
                    if(map[k][q]=='X')
                    {
                        s++;
                        if(s==5)//连接5个棋子 
                        {
                            printf("YES\n");
                            return 0;
                        }
                    }
                    else break;
                }
            }
		}
	}
    printf("NO\n");
    return 0;
}