10189 - Minesweeper

这题不难，根据题意逐行扫描处理就可以了。如果想到使用边沿哨兵可以大大减少处理难度

 

返回 Volume CI 索引

返回总索引

 

 

//////////////////////////////////////////////////////////////////////////// 10189 - Minesweeper// Copyright (c) 2010 by Bo-wen Feng// [e-mail]//////////////////////////////////////////////////////////////////////////#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <vector>#include <string>#include <sstream>#include <cstdio>#include <cmath>#include <limits>#include <utility>#ifndef ONLINE_JUDGE#include <fstream>std::ifstream cin("in.txt");std::ofstream cout("out.txt");// std::ofstream cout(stdout);#else#include <iostream>#endiftypedef long long llong;typedef unsigned long long ullong;typedef long double ldouble;using namespace std;#define MAX 500#define MINE 100int main(int argc, char* argv[]){int n, m;int field[MAX+2][MAX+2];string line;int count = 1;cin>> n>> m;while(n != 0 && m != 0){memset(field, 0, sizeof(int) * (MAX+2) * (MAX+2));for(int i = 1; i <= n; ++i){cin>> line;for(int j = 1; j <= m; ++j){if(line[j-1] == '*'){field[i][j] = MINE;++field[i-1][j-1];// TL++field[i-1][j ];// T++field[i-1][j+1];// TR++field[i ][j-1];// L++field[i ][j+1];// R++field[i+1][j-1];// BL++field[i+1][j ];// B++field[i+1][j+1];// BL}}}if(count > 1){cout<< '/n';}cout<< "Field #"<< count++<< ":/n";for(int i = 1; i <= n; ++i){for(int j = 1; j <= m; ++j){if(field[i][j] >= MINE)cout<< '*';elsecout<< field[i][j];}cout<< '/n';}cin>> n>> m;}return 0;}