Minesweeper

Minesweeper

时间： 1ms        内存：64M

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<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.

示例输入：

4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0

示例输出：

Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100

提示：

参考答案（内存最优[748]）：

/* There's a 5-case analytic solution but why bother? */

#include <stdio.h>

main(){
   int i,j,k,m,n,s,d;

   while (2 == scanf("%d%d",&s,&d)){
      int best = -9999999, tot;
      for (i=0;i<4096;i++){
         for (j=0;j<8;j++){
            for (tot=k=0;k<5;k++) tot += ((i>>(j+k))&1)?s:-d;
            if (tot >= 0) break;
         }
         if (j<8) continue;
         for (tot=j=0;j<12;j++) tot += ((i>>j)&1)?s:-d;
         if (tot > best) best = tot;
      }
      if (best > 0) printf("%d\n",best);
      else printf("Deficit\n");
   }
}

参考答案（时间最优[0]）：

#include <iostream>

using namespace std;

int main()
{
    char a[102][102]={0};
    int n,m,count=1;
    while(cin>>n>>m){
        if(n==0 && m==0)
            break;
        for(int i=1;i<=n;i++)   //input
            for(int j=1;j<=m;j++)
                cin>>a[i][j];
        for(int i=1;i<=n;i++)   //计算
            for(int j=1;j<=m;j++){
                if(a[i][j]=='*')
                    continue;

                a[i][j]='0';
                if(a[i-1][j]=='*')
                    ++a[i][j];
                if(a[i+1][j]=='*')
                    ++a[i][j];
                if(a[i][j+1]=='*')
                    ++a[i][j];
                if(a[i][j-1]=='*')
                    ++a[i][j];

                if(a[i-1][j-1]=='*')
                    ++a[i][j];
                if(a[i-1][j+1]=='*')
                    ++a[i][j];
                if(a[i+1][j-1]=='*')
                    ++a[i][j];
                if(a[i+1][j+1]=='*')
                    ++a[i][j];
            }

        if(count!=1)
            cout<<endl;
        cout<<"Field #"<<count++<<':'<<endl;
        for(int i=1;i<=n;i++){  //输出
            for(int j=1;j<=m;j++){
                cout<<a[i][j];
            }
            cout<<endl;
        }
    }
    return 0;
}

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