标签：poj zoj 算法 编程 c++
Mine Sweeper
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6429 |
|
Accepted: 2500 |
Description
The game Minesweeper is played on an n by n grid. In this grid are hidden m mines, each at a distinct grid location. The player repeatedly touches grid positions. If a position with a mine is touched, the mine explodes and the player loses. If a positon not
containing a mine is touched, an integer between 0 and 8 appears denoting the number of adjacent or diagonally adjacent grid positions that contain a mine. A sequence of moves in a partially played game is illustrated below.
Here, n is 8, m is 10, blank squares represent the integer 0, raised squares represent unplayed positions, and the figures resembling asterisks represent mines. The leftmost image represents the partially played game. From the first image to the second, the
player has played two moves, each time choosing a safe grid position. From the second image to the third, the player is not so lucky; he chooses a position with a mine and therefore loses. The player wins if he continues to make safe moves until only m unplayed
positions remain; these must necessarily contain the mines.
Your job is to read the information for a partially played game and to print the corresponding board.
Input
The first line of input contains a single postitive integer n <= 10. The next n lines represent the positions of the mines. Each line represents the contents of a row using n characters: a period indicates an unmined positon while an asterisk indicates a mined
position. The next n lines are each n characters long: touched positions are denoted by an x, and untouched positions by a period. The sample input corresponds to the middle figure above.
Output
Your output should represent the board, with each position filled in appropriately. Positions that have been touched and do not contain a mine should contain an integer between 0 and 8. If a mine has been touched, all positions with a mine should contain an
asterisk. All other positions should contain a period.
Sample Input
8
...**..*
......*.
....*...
........
........
.....*..
...**.*.
.....*..
xxx.....
xxxx....
xxxx....
xxxxx...
xxxxx...
xxxxx...
xxx.....
xxxxx...
Sample Output
001.....
0013....
0001....
00011...
00001...
00123...
001.....
00123...
注意：如果踩到地雷，要把所有地雷表示出来；
#include <iostream>
using namespace std;
#define MAX 12
int n;
char map[MAX][MAX],ans[MAX][MAX];
int count(int i,int j,int n){
int ans=0;
if (map[i-1][j-1]=='*' && i-1>=0 && j-1>=0)
ans++;
if (map[i-1][j]=='*' && i-1>=0)
ans++;
if (map[i-1][j+1]=='*' && i-1>=0 && j+1<n)
ans++;
if (map[i][j-1]=='*' && j-1>=0)
ans++;
if (map[i][j]=='*')
ans++;
if (map[i][j+1]=='*' && j+1<n)
ans++;
if (map[i+1][j-1]=='*' && i+1<n && j-1>=0)
ans++;
if (map[i+1][j]=='*' && i+1<n)
ans++;
if (map[i+1][j+1]=='*' && i+1<n && j+1<n)
ans++;
return ans;
}
int main(){
while (cin>>n){
for (int i=0;i<n;i++)
for (int j=0;j<n;j++)
cin>>map[i][j];
int flag=0;
for (int i=0;i<n;i++){
for (int j=0;j<n;j++){
cin>>ans[i][j];
if (map[i][j]=='*' && ans[i][j]=='x')
flag=1;
if (ans[i][j]=='x'){
int k=count(i,j,n);
ans[i][j]=k+'0';
}
}
}
for (int i=0;i<n;i++){
for (int j=0;j<n;j++){
if (flag==1 && map[i][j]=='*')
cout<<map[i][j];
else
cout<<ans[i][j];
}
cout<<endl;
}
}
return 0;
}
poj 2612 Mine Sweeper
标签：poj zoj 算法 编程 c++
原文地址：http://blog.csdn.net/codeforcer/article/details/41289131
