1096: Minesweeper扫雷
范翰海
题目原型
题目描述
问题 1096: Minesweeper
时间限制: 1Sec 内存限制: 64MB 提交: 2393 解决: 981
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 1*10 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
1*10
1110
Field #2:
**100
33200
1*100
提示
无
代码
#include<iostream>
using namespace std;
int main()
{
int row,column,first=0;
while(cin>>row>>column){
first++;
if(row==0&&column==0)
break;
else{
char aa[row+1][column+1];
for(int i=0;i<row;i++){
for(int j=0;j<column;j++){
cin>>aa[i][j];//cout<<111111<<endl;
}
}
cout<<"Field #"<<first<<":"<<endl;
char aaa[row][column];
for(int i=0;i<row;i++){
for(int j=0;j<column;j++){
if(aa[i][j]=='*')
{
cout<<'*';continue;
}
int round=0;
if(i>0&&j>0&&i<row-1&&j<column-1){
if(aa[i-1][j-1]=='*')round++;
if(aa[i-1][j]=='*')round++;
if(aa[i-1][j+1]=='*')round++;
if(aa[i][j-1]=='*')round++;
if(aa[i][j+1]=='*')round++;
if(aa[i+1][j-1]=='*')round++;
if(aa[i+1][j]=='*')round++;
if(aa[i+1][j+1]=='*')round++;
}
else if(i==0&&j!=0&&j!=column-1){
// cout<<123123<<endl;
if(aa[i][j-1]=='*')round++;
if(aa[i][j+1]=='*')round++;
if(aa[i+1][j-1]=='*')round++;
if(aa[i+1][j]=='*')round++;
if(aa[i+1][j+1]=='*')round++;
}
else if(i!=0&&i!=row-1&&j==0){
if(aa[i-1][j]=='*')round++;
if(aa[i+1][j]=='*')round++;
if(aa[i-1][j+1]=='*')round++;
if(aa[i][j+1]=='*')round++;
if(aa[i+1][j+1]=='*')round++;
}
else if(i==row-1&&j!=0&&j!=column-1){
if(aa[i-1][j]=='*')round++;
if(aa[i-1][j-1]=='*')round++;
if(aa[i-1][j+1]=='*')round++;
if(aa[i][j-1]=='*')round++;
if(aa[i][j+1]=='*')round++;
}
else if(i!=0&&i!=row-1&&j==column-1){
if(aa[i-1][j]=='*')round++;
if(aa[i+1][j]=='*')round++;
if(aa[i-1][j-1]=='*')round++;
if(aa[i][j-1]=='*')round++;
if(aa[i+1][j-1]=='*')round++;
}
else if(i==0&&j==0){
if(aa[i][j+1]=='*')round++;
if(aa[i+1][j]=='*')round++;
if(aa[i+1][j+1]=='*')round++;
}
else if(i==0&&j==column-1){
if(aa[i][j-1]=='*')round++;
if(aa[i+1][j]=='*')round++;
if(aa[i+1][j-1]=='*')round++;
}
else if(j==0&&i==row-1){
if(aa[i-1][j]=='*')round++;
if(aa[i-1][j+1]=='*')round++;
if(aa[i][j+1]=='*')round++;
}
else if(j==column-1&&i==row-1){
if(aa[i-1][j]=='*')round++;
if(aa[i-1][j-1]=='*')round++;
if(aa[i][j-1]=='*')round++;
}
cout<<round;
}
cout<<endl;
}
cout<<endl;
}
}
return 0;
}
总结
1，只要思想不滑坡，办法总比困难多，
2，这题很简单，但是要分情况讨论，我的代码分了9总情况，可以根据题目中给的例子来分析这9种情况的代码应该怎么写
3，一定要注意细节，要不然真的卡的很难受eg，两层for循环中第一层为i，第二层为j，一定不要打乱
4，总结一个小tip，单个字符用int a,数字用aa[]，来表示容易区分，最好能顾名思义，英语优先，汉语也比随便写强
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