描述
I
think most of you are using system named of xp or vista or win7.And these system
is consist of a famous game what is mine sweeping.You must have played it
before.If you not,just look the game rules followed.
There are N*N grids
on the map which contains some mines , and if you touch that ,you lose the
game.If a position not containing a mine is touched, an integer K (0 < =K
<= 8) appears indicating that there are K mines in the eight adjacent
positions. If K = 0, the eight adjacent positions will be touched
automatically, new numbers will appear and this process is repeated until no
new number is 0. Your task is to mark the mines‘ positions without touching
them.
Now, given the distribution of the mines, output the numbers
appearing after the player‘s first touch.
输入
The first line of each case is two numbers N (1 <= N <= 100) .Then there will be a map contain N*N grids.The map is just contain O and X.‘X‘ stands for a mine, ‘O‘ stand for it is safe with nothing. You can assume there is at most one mine in one position. The last line of each case is two numbers X and Y(0<=X<N,0<=Y<N, indicating the position of the player‘s first touch.
输出
If the player touches
the mine, just output "it is a beiju!".
If the player doesn‘t touch the
mine, output the numbers appearing after the touch. If a position is touched by
the player or by the computer automatically, output the number. If a position
is not touched, output a dot ‘.‘.
Output a blank line after each test
case.
样例输入
5
OOOOO
OXXXO
OOOOO
OXXXO
OOOOO
1 1
5
OOOOO
OXXXO
OOOOO
OXXXO
OOOOO
0 0
样例输出
it is a beiju!
1....
.....
.....
.....
.....
题目来源
扫雷游戏~
#include <stdio.h>#include <string.h>#define MAXN 150int
N;int
flag[MAXN][MAXN];char
map[MAXN][MAXN];char
out[MAXN][MAXN];int
judge(int
x, int
y){ int
flag=0; for(int
i=x-1; i<=x+1; i++){ for(int
j=y-1; j<=y+1; j++){ if(1<=i && i<=N && 1<=j && j<=N && !(i==x&&j==y)){ if(map[i][j]==‘X‘)flag++; } } } return
flag;}void
dfs(int
x, int
y){ int
sum=judge(x,y); if(sum==0){ out[x][y]=‘0‘; for(int
i=x-1; i<=x+1; i++){ for(int
j=y-1; j<=y+1; j++){ if(1<=i && i<=N && 1<=j && j<=N && !(i==x&&j==y)){ if(!flag[i][j]) dfs(i,j); } flag[i][j]=1; } } }else{ out[x][y]=sum+‘0‘; }}int
main(){ while( scanf("%d",&N)!=EOF ){ for(int
i=1; i<=N; i++){ getchar(); for(int
j=1; j<=N; j++){ scanf("%c",&map[i][j]); } } int
x,y; scanf("%d %d",&x,&y); x++; y++; if(map[x][y]==‘X‘){ puts("it is a beiju!\n"); continue; } memset(flag,0,sizeof(flag)); memset(out,‘.‘,sizeof(out)); dfs(x,y); for(int
i=1; i<=N; i++){ for(int
j=1; j<=N; j++){ printf("%c",out[i][j]); } printf("\n"); } printf("\n"); } return
0;} |
原文：http://www.cnblogs.com/chenjianxiang/p/3539689.html