G - Five-In-a-Row CodeForces - 825B

Alice and Bob play 5-in-a-row game. They have a playing field of size10 × 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'.

Example

Input

XX.XX.....
.....OOOO.
..........
..........
..........
..........
..........
..........
..........
..........

Output

YES

Input

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

Output

NO

给你一个棋盘，上面已经有四个黑子和四个白子，问再下一个黑子黑色的能不能连成五个，
我的思路就是去找黑色的棋子，因为棋子必须在一条线上，所以直接设置方向搜索即可，及向上，向下，向左，向右，左上，左下，右上，右下，

因为会出现XX。XX的情况，所以设置一个变量去记录。的个数，假如超过了1个则说明黑子不能连成五个。
#include<stdio.h>
#include<string.h>
char map1[20][20];//存放棋盘
int main()
{
    while(~scanf("%s",map1[0]))//输入棋盘
    {
        int flag=0;
        for(int i=1; i<10; i++)
            scanf("%s",map1[i]);
        for(int i=0; i<10; i++)
        {
            for(int j=0; j<10; j++)
            {
                if(map1[i][j]=='X')//当找到棋子x时，对周围进行搜索
                {
                    int gs=1,d=0;//gs代表找到的棋子数，因为有xx.xx等这样的情况存在，所以设置一个变量d，使得黑棋与黑棋之间可以隔一个
                    for(int k=j+1; k<10; k++)//发现以后向左寻找
                    {
                        if(map1[i][k]=='O')//如果遇到白棋则说明这个方向上是肯定组不成五个了所以跳出这个方向
                            break;
                        else if(map1[i][k]=='.')//如果是.则点可以把黑棋分成两半，所以用d来记录点的个数
                        {
                            d++;
                            if(d>1)//如果点多于一个则肯定连不上
                                break;
                            gs++;//因为要在点处放旗子，所以也要记录，相当于在那里放了一个黑棋子
                        }
                       else if(map1[i][k]=='X')//找到棋子，gs++；
                            gs++;
                        if(gs==5)//当找到的棋子数为伍时，则输出yes，退出程序。
                        {
                            printf("YES\n");flag=1; return 0;
                        }
                    }
                    gs=1,d=0;//每次寻找都把棋子数和点数重置
                    for(int k=j-1; k>=0; k--)//向右寻找
                    {
                        if(map1[i][k]=='O')
                            break;
                        else if(map1[i][k]=='.')
                        {
                            d++;
                            if(d>1)break;
                            gs++;
                        }
                        else if(map1[i][k]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n"); flag=1;return 0;
                        }
                    }
                    gs=1,d=0;
                    for(int k=i+1; k<10; k++)//向下寻找
                    {
                        if(map1[k][j]=='O')
                            break;
                        else if(map1[k][j]=='.')
                        {
                            d++;
                            if(d>1)break;
                            gs++;
                        }
                        else if(map1[k][j]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n"); flag=1;return 0;
                        }
                    }
                    gs=1,d=0;
                    for(int k=i-1; k>=0; k--)//向上寻找
                    {
                        if(map1[k][j]=='O')
                            break;
                        else if(map1[k][j]=='.')
                        {
                            d++;
                            if(d>1)break;
                            gs++;
                        }
                        else if(map1[k][j]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n"); flag=1; return 0;
                        }
                    }
                    gs=1,d=0;
                    for(int k=j+1,q=i+1; k<10&&q<10; k++,q++)//向右下寻找
                    {
                        if(map1[q][k]=='O')
                            break;
                        else if(map1[q][k]=='.')
                        {
                            d++;
                            if(d>1)break;
                            gs++;
                        }
                        else if(map1[q][k]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n"); flag=1;return 0;  
                        }
                    }
                    gs=1,d=0;
                    for(int k=j-1,q=i-1; k>=0&&q>=0; k--,q--)//向左上寻找
                    {
                        if(map1[q][k]=='O')
                            break;
                        else if(map1[q][k]=='.')
                        {
                            d++;
                            if(d>1)break;
                            gs++;
                        }
                        else if(map1[q][k]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n");flag=1;return 0;
                        }
                    }
                    gs=1,d=0;
                    for(int k=j-1,q=i+1; k>=0&&q<10; k--,q++)//向左下寻找
                    {
                        if(map1[q][k]=='O')
                            break;
                        else if(map1[q][k]=='.')
                        {
                            d++;
                            if(d>1)break;  
                            gs++;
                        }
                        else if(map1[q][k]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n"); flag=1; return 0;
                        }
                    }
                    gs=1,d=0;
                    for(int k=j+1,q=i-1; k<=10&&q>=0; k++,q--)//向右上寻找
                    {
                        if(map1[q][k]=='O')
                            break;
                        else if(map1[q][k]=='.')
                        {
                            d++;
                            if(d>1)break;
                            gs++;
                        }
                        else if(map1[q][k]=='X')
                            gs++;
                        if(gs==5)
                        {
                            printf("YES\n");flag=1;return 0;
                        }
                    }
                }
            }
        }
        if(flag==0)//若flag==0，则说明没有找到合适的地方，所以输出no
            printf("NO\n");
    }
    return 0;
}