CodeForces - 825B Five-In-a-Row

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元素不会背元素周期表

最新推荐文章于 2024-07-31 17:49:42 发布

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本文介绍了一种通过编程方式来判断五子棋游戏中玩家是否能在下一步获胜的方法。通过遍历棋盘上的每个空位，并检查该位置周围的棋子状态，可以确定是否存在使玩家立即获胜的走法。

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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'.

Example
Input
XX.XX.....
.....OOOO.
..........
..........
..........
..........
..........
..........
..........
..........
Output
YES
Input
XXOXX.....
OO.O......
..........
..........
..........
..........
..........
..........
..........
..........
Output
NO

题意：五子棋，找出横排或者竖排或者对角线能组成五个'X'则输出“YES”，否则输出“NO”。

用暴力，找出一个' . ' 点，从它的横排。竖排，和对角线开始搜，如果和它相邻的'X'数量>=4，就能组成五子棋，用f标记已经找到，break。

若f=1，输出“YES”,否则输出“NO”。

#include<stdio.h>
#include<string.h>
int main()
{
    char str[20][20];
    int f;
    while(~scanf("%s",str[0]))
    {
        for(int i=1; i<10; i++)
            scanf("%s",str[i]);
        f=0;
        int sum;
        for(int i=0; i<10; i++)
        {
            if(f)
                break;
            for(int j=0; j<10; j++)
            {
                if(str[i][j]=='.')
                {
                    sum=0;
                    for(int k=i+1; k<10; k++)//竖排向下搜
                        if(str[k][j]=='X')
                            sum++;
                        else
                            break;
                    for(int k=i-1; k>=0; k--)//竖排向上搜
                        if(str[k][j]=='X')
                            sum++;
                        else
                            break;
                    if(sum>=4)
                    {
                        f=1;
                        break;
                    }

                    sum=0;
                    for(int k=j+1; k<10; k++)  横排向右搜
                        if(str[i][k]=='X')
                            sum++;
                        else
                            break;
                    for(int k=j-1; k>=0; k--)横排向左搜
                        if(str[i][k]=='X')
                            sum++;
                        else
                            break;
                    if(sum>=4)
                    {
                        f=1;
                        break;
                    }

                    sum=0;
                    for(int k=i+1,r=j+1; k<10&& r<10; k++,r++)右下
                        if(str[k][r]=='X')
                            sum++;
                        else
                            break;
                    for(int k=i-1,r=j-1; k>=0&&r>=0; k--,r--)左上
                        if(str[k][r]=='X')
                            sum++;
                        else
                            break;
                    if(sum>=4)
                    {
                        f=1;
                        break;
                    }
                    sum=0;
                    for(int k=i+1,r=j-1; k<10&&r>=0; k++,r--)//左下
                        if(str[k][r]=='X')
                            sum++;
                        else
                            break;
                    for(int k=i-1,r=j+1; k>=0&&r<10; k--,r++)//右上
                        if(str[k][r]=='X')
                            sum++;
                        else
                            break;
                    if(sum>=4)
                    {
                        f=1;
                        break;
                    }
                }

            }
        }
        if(f)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}