题目 1096: Minesweeper＜扫雷＞

最新推荐文章于 2022-04-03 19:26:23 发布

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最新推荐文章于 2022-04-03 19:26:23 发布

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分类专栏： C/C++ 文章标签： c语言算法

本文链接：https://blog.youkuaiyun.com/qq_44800871/article/details/112739984

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时间限制: 1Sec 内存限制: 64MB 提交: 3911 解决: 1644

题目描述

Minesweeper Have you ever played Minesweeper? This cute little game comes with a certain operating system whose name we can’t remember. The goal of the game is to find where all the mines are located within a M x N field. The game shows a number in a square which tells you how many mines there are adjacent to that square. Each square has at most eight adjacent squares. The 4 x 4 field on the left contains two mines, each represented by a ``*’’ character. If we represent the same field by the hint numbers described above, we end up with the field on the right: … … .… … 100 2210 110 1110

输入

The input will consist of an arbitrary number of fields. The first line of each field contains two integers n and m ( 0 < n, m$ \le$100) which stand for the number of lines and columns of the field, respectively. Each of the next n lines contains exactly m characters, representing the field. Safe squares are denoted by ‘.’ and mine squares by ‘*’, both without the quotes. The first field line where n = m = 0 represents the end of input and should not be processed.

输出

For each field, print the message Field #x: on a line alone, where x stands for the number of the field starting from 1. The next n lines should contain the field with the ``.’’ characters replaced by the number of mines adjacent to that square. There must be an empty line between field outputs.

样例输入

4 4
…
…
.…
…
3 5
**…
…
.*…
0 0

样例输出

Field #1:
100
2210
110
1110

Field #2:
**100
33200
1*100

程序实现

#include<stdio.h>
char field[200][200];//地图 
int calculate[200][200];//探测数组 
int main()
{
	int n,m;//行数和列数
	int time=0;//记录地雷在附近出现的个数 
	while(scanf("%d %d",&n,&m)&&n!=0&&m!=0)
	{
		time++;
		for(int i=0;i<n;i++)
		{
			scanf("%s",field[i]);//注意二维字符数组的输入方式，不可用%c按字符读入 
			for(int j=0;j<m;j++)
			{
				calculate[i][j]=0;
			}
		}
		printf("Field #%d:\n",time);
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(field[i][j]=='*')
				{
					//calculate[i][j]=-1;
					printf("*");//地雷 
				}
				else
				{	//探测该方块的八个方位，出现地雷则该位置探测数加一 
					if(i-1>=0&&j-1>=0&&field[i-1][j-1]=='*')
						calculate[i][j]++;
					if(i-1>=0&&field[i-1][j]=='*')
						calculate[i][j]++;
					if(i-1>=0&&j+1<m&&field[i-1][j+1]=='*')
						calculate[i][j]++;
					if(j-1>=0&&field[i][j-1]=='*')
						calculate[i][j]++;
					if(j+1<m&&field[i][j+1]=='*')
						calculate[i][j]++;
					if(i+1<n&&j-1>=0&&field[i+1][j-1]=='*')
						calculate[i][j]++;
					if(i+1<n&&field[i+1][j]=='*')
						calculate[i][j]++;
					if(i+1<n&&j+1<n&&field[i+1][j+1]=='*')
						calculate[i][j]++;
					printf("%d",calculate[i][j]);//打印探测数 
				}
			}
			printf("\n");
		}
		printf("\n");
	}
	return 0;
}