问题 1096: Minesweeper
时间限制: 1Sec 内存限制: 64MB 提交: 581 解决: 254
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
#include<cstdio> #include<iostream> #include<vector> #include<string.h> using namespace std; char pic[1001][1001]; int vis[1001][1001]; int dx[]={0,0,1,-1,1,-1,1,-1}; int dy[]={1,-1,0,0,1,-1,-1,1}; int h,w; struct node{ int i; int j; }; vector<node>v;//雷 vector<node>l;//空白 void print(int i,int j){//扫描雷 for(int k=0;k<8;k++){ int tmpi=i+dy[k],tmpj=j+dx[k]; if(tmpi>=0 && tmpi<h && tmpj>=0 && tmpj<w) vis[tmpi][tmpj]++; } return; } void printb(int i,int j){//扫描空白 int ans=0; for(int k=0;k<8;k++){ int tmpi=i+dy[k],tmpj=j+dx[k]; if(tmpi>=0 && tmpi<h && tmpj>=0 && tmpj<w && pic[tmpi][tmpj]=='*') ans++; } vis[i][j]=ans; return; } int main(){ int tot=1; while(cin>>h>>w && (h+w)!=0){ memset(vis,0,sizeof(vis)); v.clear(); l.clear(); int tota=0,totb=0;//a雷b空白初始化 for(int i=0;i<h;i++){ for(int j=0;j<w;j++){ cin>>pic[i][j]; if(pic[i][j]=='*') {node cc;cc.i=i;cc.j=j;v.push_back(cc);tota++;} if(pic[i][j]=='.') {node cc;cc.i=i;cc.j=j;l.push_back(cc);totb++;} } } printf("Field #%d:\n",tot++); if(tota>totb){ for(int i=0;i<l.size();i++){ printb(l[i].i,l[i].j); } }else{ for(int i=0;i<v.size();i++){ print(v[i].i,v[i].j); } } for(int i=0;i<h;i++){ for(int j=0;j<w;j++){ if(pic[i][j]=='*') printf("*"); else printf("%d",vis[i][j]); } printf("\n"); } printf("\n"); } return 0; }