CodeforcesRound#483(Div.2)B.Minesweeper
B. Minesweeper
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
One day Alex decided to remember childhood when computers were not too powerful and lots of people played only default games. Alex enjoyed playing Minesweeper that time. He imagined that he saved world from bombs planted by terrorists, but he rarely won.
Alex has grown up since then, so he easily wins the most difficult levels. This quickly bored him, and he thought: what if the computer gave him invalid fields in the childhood and Alex could not win because of it?
He needs your help to check it.
A Minesweeper field is a rectanglen×mn×m, where each cell is either empty, or contains a digit from11to88, or a bomb. The field is valid if for each cell:
- if there is a digitkkin the cell, then exactlykkneighboring cells have bombs.
- if the cell is empty, then all neighboring cells have no bombs.
Two cells are neighbors if they have a common side or a corner (i.e. a cell has at most88neighboring cells).
Input
The first line contains two integersnnandmm(1≤n,m≤1001≤n,m≤100) — the sizes of the field.
The nextnnlines contain the description of the field. Each line containsmmcharacters, each of them is "." (if this cell is empty), "*" (if there is bomb in this cell), or a digit from11to88, inclusive.
Output
Print "YES", if the field is valid and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrarily.
Examples
input
Copy
3 3 111 1*1 111
output
Copy
YES
input
Copy
2 4 *.*. 1211
output
Copy
NO
Note
In the second example the answer is "NO" because, if the positions of the bombs are preserved, the first line of the field should be*2*1.
You can read more about Minesweeper inWikipedia's article.
给你一个n*m的矩阵，问是否合理。其中*表示炸弹，‘.’表示这一点附近八个方向都没有炸弹，其他数字表示这一点附近八个方向炸弹个数。
也就是给你的矩阵，‘*’是已知，判断别的点对不对。
我们处理字符矩阵，把是'*'的附近八个方向都加加放在另外一个数组中，最后对比原来的数组就可以。
#include <iostream> #include <algorithm> #include <cstring> using namespace std; int mp[105][105]; int ans[105][105]; int n,m; int dir[][2]={-1,1, 0,1, 1,1, -1,0, 1,0, -1,-1, 0,-1, 1,-1}; int main() { while(~scanf("%d %d",&n,&m)) { memset(mp,-1,sizeof(mp)); memset(ans,0,sizeof(ans)); char tp; getchar(); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { scanf("%c",&tp); if(tp=='*') {mp[i][j]=-2;//-2表示有炸弹 for(int k=0;k<8;k++) { int tx,ty; tx=dir[k][0]+i; ty=dir[k][1]+j; if(i>=1&&i<=n&&j>=1&&j<=m&&mp[i][j]!=-1) ans[tx][ty]++; } } else if(tp=='.') mp[i][j]=0,ans[i][j]=0;//-1表示空 else mp[i][j]=tp-'0'; if(j==m)getchar(); } int flag=1; for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { if(mp[i][j]!=-1&&mp[i][j]!=-2&&mp[i][j]!=ans[i][j]) {//printf("%d %d %d %d",ans[i][j],mp[i][j],i,j); flag=0; break; } } if(flag) printf("YES "); else printf("NO "); } return 0; }
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