Codeforces Round #483 (Div. 2) B题
1 second
256 megabytes
standard input
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 rectangle n×m
, where each cell is either empty, or contains a digit from 1 to 8
, or a bomb. The field is valid if for each cell:
- if there is a digit k
in the cell, then exactly k
- neighboring 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 most 8
neighboring cells).
The first line contains two integers n
and m (1≤n,m≤100
) — the sizes of the field.
The next n
lines contain the description of the field. Each line contains m characters, each of them is "." (if this cell is empty), "*" (if there is bomb in this cell), or a digit from 1 to 8
, inclusive.
Print "YES", if the field is valid and "NO" otherwise.
You can choose the case (lower or upper) for each letter arbitrarily.
3 3
111
1*1
111
YES
2 4
*.*.
1211
NO
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 in Wikipedia's article.
思路:简单搜索
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <vector>
//codeforces
using namespace std;
char maps[110][110];
int dir[8][2]={{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};
int n,m;
int dfs(int x,int y)
{
int tx,ty;
if(maps[x][y]=='.'){
for(int i=0;i<8;i++){
tx=x+dir[i][0];
ty=y+dir[i][1];
if(tx<0||tx>=n||ty<0||ty>=m) continue;
if(maps[tx][ty]=='*'){
return 1;
}
}
}
int tmpe,countt=0;
if(maps[x][y]<='8'&&maps[x][y]>='1'){
tmpe=maps[x][y]-'0';
for(int i=0;i<8;i++){
tx=x+dir[i][0];
ty=y+dir[i][1];
if(tx<0||tx>=n||ty<0||ty>=m) continue;
if(maps[tx][ty]=='*'){
countt++;
}
}
if(countt!=tmpe) return 1;
}
return 0;
}
int main()
{
int flag;
while(scanf("%d %d",&n,&m)!=EOF){
flag=0;
for(int i=0;i<n;i++){
scanf("%s",maps[i]);
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(maps[i][j]=='.'){
if(dfs(i,j)){
flag=1;
break ;
}
}
if(maps[i][j]<='8'&&maps[i][j]>='1'){
if(dfs(i,j)){
flag=1;break;
}
}
}
}
if(flag) printf("NO\n");
else printf("YES\n");
}
return 0;
}
Codeforces Round #483 (Div. 2) B题的更多相关文章
- Codeforces Round #483 (Div. 2)C题
C. Finite or not? time limit per test 1 second memory limit per test 256 megabytes input standard in ...
- Codeforces Round #378 (Div. 2) D题(data structure)解题报告
题目地址 先简单的总结一下这次CF,前两道题非常的水,可是第一题又是因为自己想的不够周到而被Hack了一次(或许也应该感谢这个hack我的人,使我没有最后在赛后测试中WA).做到C题时看到题目情况非常 ...
- Codeforces Round #612 (Div. 2) 前四题题解
这场比赛的出题人挺有意思,全部magic成了青色. 还有题目中的图片特别有趣. 晚上没打,开virtual contest打的,就会前三道,我太菜了. 最后看着题解补了第四道. 比赛传送门 A. An ...
- Codeforces Round #713 (Div. 3)AB题
Codeforces Round #713 (Div. 3) Editorial 记录一下自己写的前二题本人比较菜 A. Spy Detected! You are given an array a ...
- Codeforces Round #552 (Div. 3) A题
题目网址:http://codeforces.com/contest/1154/problem/ 题目意思:就是给你四个数,这四个数是a+b,a+c,b+c,a+b+c,次序未知要反求出a,b,c,d ...
- Codeforces Round #412 Div. 2 补题 D. Dynamic Problem Scoring
D. Dynamic Problem Scoring time limit per test 2 seconds memory limit per test 256 megabytes input s ...
- Codeforces Round #271 (Div. 2) E题 Pillars(线段树维护DP)
题目地址:http://codeforces.com/contest/474/problem/E 第一次遇到这样的用线段树来维护DP的题目.ASC中也遇到过,当时也非常自然的想到了线段树维护DP,可是 ...
- Codeforces Round #425 (Div. 2))——A题&&B题&&D题
A. Sasha and Sticks 题目链接:http://codeforces.com/contest/832/problem/A 题目意思:n个棍,双方每次取k个,取得多次数的人获胜,Sash ...
- Codeforces Round #483 (Div. 2) [Thanks, Botan Investments and Victor Shaburov!]
题目链接:http://codeforces.com/contest/984 A. Game time limit per test:2 seconds memory limit per test:5 ...
随机推荐
- error C3861: “getpid”: 找不到标识符
原文:http://blog.csdn.net/woniu199166/article/details/52471242 这种错误一般就是没有对应的函数或者对应的头文件 旧版的vs添加#include ...
- mysql存储过程模板
CREATE DEFINER=`root`@`localhost` PROCEDURE `SP_test`(IN `nodeCode` varchar(100),IN `id` varchar(36) ...
- 《深入理解JVM》第二章读书笔记
Java内存区域与内存溢出异常 运行时数据区域 JVM执行java程序的时候有一个运行时数据区,每个区域有自己的作用,了解这些区域有助于我们理解JVM.JVM运行时数据区如图所示: 程序计数器 该区域 ...
- ionic项目编译打包(android平台)
ionic项目相关开发工作完成之后(建立ionic工程项目可以参考上一篇文章ionic项目工程建立),就可以进行项目的编译打包apk应用包. 打包编译需要在平台环境下,这里只记录下android平台打 ...
- C# 日期和时间的字符串表示形式转换为其等效的DateTime(stringToDateTime)
一. 标准的日期和时间字符串转换 将日期和时间的字符串表示形式转换为其等效的DateTime对象是开发中很常见的类型转换,我们最常使用的方式是: // 如果s为null,抛出ArgumentNullE ...
- [翻译] IGLDropDownMenu
IGLDropDownMenu An iOS drop down menu with pretty animation. 一种iOS点击下拉菜单样式,动画效果很绚丽. Screenshot - 截图 ...
- Java学习---程序设计_面试题[2]
百度2017春招笔试真题编程题集合之买帽子 // 2017-10-09 // 题目描述 // 度度熊想去商场买一顶帽子,商场里有N顶帽子,有些帽子的价格可能相同.度度熊想买一顶价格第三便宜的帽子,问第 ...
- pandas 入门
1. 默认的pandas不能读取excel.需要安装xlrd, xlwt才能支持excel的读写 pip install xlrd #添加读取excel功能 pip install xlwt #添加写 ...
- ZOJ-3278 8G Island---二分第k大
题目链接: https://cn.vjudge.net/problem/ZOJ-3278 题目大意: 给出两个数列A和B,长度分别为N,M (1<=N, M<=10^5, 1<=Ai ...
- [19/04/24-星期三] GOF23_创建型模式(建造者模式、原型模式)
一.建造者模式 本质:分离了对象子组件的单独构造(由Builder负责)和装配的分离(由Director负责),从而可以构建出复杂的对象,这个模式适用于:某个对象的构建过程十分复杂 好处:由于构建和装 ...