ZOJ - 1628 Diamond

本文介绍了一款在8*8棋盘上进行的钻石游戏,玩家通过交换相邻钻石的位置来形成三个及以上同色钻石的连线,实现消除。文章提供了判断玩家操作是否合法的算法实现,包括输入坐标验证和消除条件检查。

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Time Limit: 2MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Status

Description

Diamond mine is a mini-game which is played on an 8 * 8 board as you can see below.

The board is filled with different colors of diamonds. The player can make one move at a time. A move is legal if it swaps two adjacent diamonds(not diagonally) and after that, there are three or more adjacent diamonds in a row or column with the same color. Those diamonds will be taken away and new diamonds will be put in their positions. The game continues until no legal moves exist.


Given the board description. You are going to determine whether a move is legal.


Input

The input contains several cases. Each case has exactly 9 lines. The first 8 lines each contains a string of 8 characters. The characters are 'R'(Red), 'O'(Orange), 'G'(Green), 'P'(Purple), 'W'(White), 'Y'(Yellow) or 'B'(Blue). All characters are uppercase. No 3 diamonds of the same color are initially in adjacent positions in a row or column. The last line has 4 integers in the form " row1 column1 row2 column2" describing the postions of the 2 diamonds that the player wants to swap. Rows are marked 1 to 8 increasingly from top to bottom while columns from left to right. Input is terminated by EOF.


Output

For each case, output "Ok!" if the move is legal or "Illegal move!" if it is not.


Sample Input

PBPOWBGW
RRPRYWWP
YGBYYGPP
OWYGGRWB
GBBGBGGR
GBWPPORG
PPGORWOG
WYWGYWBY
4 3 3 3
PBPOWBGW
RRPRYWWP
YGBYYGPP
OWYGGRWB
GBBGBGGR
GBWPPORG
PPGORWOG
WYWGYWBY
5 5 6 5


Sample Output

Ok!
Illegal move!




这题看似简单其实特别坑输入的坐标还有可能不相邻和超过范围。。。真是无节操坑


#include <iostream>

#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 20;
char str[N][N];
int judge(int x,int y);


int main()
{
    while(scanf("%s",str[0])!=EOF)
    {
        for(int i=1; i<8; i++)
        {
            scanf(" %s",str[i]);
        }
        int x1, y1, x2, y2;
        scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
        int flag=0;
        if((abs(x1-x2)+abs(y2-y1)!=1)||(x1<1)||(x1>8)||(x2<1)||(x2>8)||(y1<1)||(y1>8)||(y2<1)||(y2>8))//技巧
        {
            flag=0;
        }
        else
        {
            char tmp;
            tmp=str[x1-1][y1-1],str[x1-1][y1-1]=str[x2-1][y2-1],str[x2-1][y2-1]=tmp;
            for(int i=0; i<8; i++)
            {
                for(int j=0; j<8; j++)
                {
                    if(judge(i,j))
                    {
                        flag=1;
                        break;
                    }
                }
                if(flag==1)
                {
                    break;
                }
            }
        }
        if(flag==1)
        {
            printf("Ok!\n");
        }
        else
        {
            printf("Illegal move!\n");
        }
        memset(str,0,sizeof(str));
    }
    return 0;
}


int judge(int x,int y)
{
    if(y<6&&str[x][y+1]==str[x][y+2]&&str[x][y+1]==str[x][y])
    {
        return 1;
    }
    if(x<6&&str[x+1][y]==str[x+2][y]&&str[x+1][y]==str[x][y])
    {
        return 1;
    }
    return 0;
}
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