Problem B: Bulbs

Greg has an m×n grid of Sweet Lightbulbs of Pure Coolness he would like to turn on. Initially, some of the bulbs are on and some are off. Greg can toggle some bulbs by shooting his laser at them. When he shoots his laser at a bulb, it toggles that bulb between on and off. But, it also toggles every bulb directly below it, and every bulb directly to the left of it. What is the smallest number of times that Greg needs to shoot his laser to turn all the bulbs on?

Input

The first line of input contains a single integer T (1 ≤ T ≤ 10), the number of test cases. Each test case starts with a line containing two space-separated integers m and n (1 ≤ m,n ≤ 400). The next m lines each consist of a string of length n of 1s and 0s. A 1 indicates a bulb which is on, and a 0 represents a bulb which is off.

Output

For each test case, output a single line containing the minimum number of times Greg has to shoot his laser to turn on all the bulbs.
Sample Input

2

3 4

0000

1110

1110

2 2

10

00

Sample Output

1

2
Explanation
In the first test case, shooting a laser at the top right bulb turns on all the bulbs which are off, and does not toggle any bulbs which are on.
In the second test case, shooting the top left and top right bulbs will do the job.

题意:把所有0,变成1最少需要几次;规则:如果把0变成1同时把左边的数和下边的数同时改变,

题解:从左上角开始处理,for循环遍历即可

#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
using namespace std;
int n,m,ans=;
int a[],b[];
string s[];
int main()
{
int t;
cin>>t;
while(t--)
{
cin>>n>>m;
ans=;
memset(a,,sizeof(a));
memset(b,,sizeof(b));
for(int i=;i<n;i++)
cin>>s[i];
for(int i=;i<n;i++)
{
for(int j=m-;j>=;j--)
{
int x=s[i][j]-''+a[i]+b[j];
if(x%==)
{
ans++;
a[i]++;
b[j]++;
}
}
}
cout<<ans<<endl;
}
}

Problem B: Bulbs的更多相关文章

  1. Codeforces Round #338 (Div. 2) A. Bulbs 水题

    A. Bulbs 题目连接: http://www.codeforces.com/contest/615/problem/A Description Vasya wants to turn on Ch ...

  2. HDU 5601 N*M bulbs 找规律

    N*M bulbs 题目连接: http://codeforces.com/contest/510/problem/C Description NM个灯泡排成一片,也就是排成一个NM的矩形,有些开着, ...

  3. BestCoder Round #67 (div.2) N bulbs(hdu 5600)

    N bulbs Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Su ...

  4. N bulbs(规律)

    N bulbs  Accepts: 408  Submissions: 1224  Time Limit: 10000/5000 MS (Java/Others)  Memory Limit: 655 ...

  5. zoj 2976 Light Bulbs(暴力枚举)

    Light Bulbs Time Limit: 2 Seconds      Memory Limit: 65536 KB Wildleopard had fallen in love with hi ...

  6. 哈理工2015 暑假训练赛 zoj 2976 Light Bulbs

    MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu SubmitStatusid=14946">Practice ...

  7. 1199 Problem B: 大小关系

    求有限集传递闭包的 Floyd Warshall 算法(矩阵实现) 其实就三重循环.zzuoj 1199 题 链接 http://acm.zzu.edu.cn:8000/problem.php?id= ...

  8. No-args constructor for class X does not exist. Register an InstanceCreator with Gson for this type to fix this problem.

    Gson解析JSON字符串时出现了下面的错误: No-args constructor for class X does not exist. Register an InstanceCreator ...

  9. C - NP-Hard Problem(二分图判定-染色法)

    C - NP-Hard Problem Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:262144 ...

随机推荐

  1. Python数据类型-7 bytes

    bytes 在Python3以后,字符串和bytes类型彻底分开了.字符串是以字符为单位进行处理的,bytes类型是以字节为单位处理的. bytes数据类型在所有的操作和使用甚至内置方法上和字符串数据 ...

  2. STM32的程序升级

    IAP基础参考http://www.eeworld.com.cn/mcu/2018/ic-news112042038.html https://blog.csdn.net/tq384998430/ar ...

  3. GsonUtils.getGson().fromJson() 转泛型集合用法

    //计算其他收费 List<QiTaFree> qiTaFreeList = GsonUtils.getGson().fromJson(exhiMain.getQiTaFressJson( ...

  4. 调用百度汇率api 获取各国的汇率值

    设置一个定时任务,每天更新汇率java代码如下 package com.thinkgem.jeesite.modules.huiLvApi.service; import java.io.Buffer ...

  5. js获取一个页面 是从哪个页面过来的

    document.referrer 获取来源页面的url console.log(document.referrer) if(document.referrer=="http://127.0 ...

  6. java 调用阿里云短信接口,报InvalidTimeStamp.Expired : Specified time stamp or date value is expired.

    官网解释: 问题所在: 自己的电脑(或者服务器) 的时间与阿里云的服务器时间 相差15分钟了. 解决方法 : 把自己的电脑时间 (或者服务器)的时间 改成标准的北京时间就行了.

  7. 微信红包系统设计 & 优化

    微信红包系统设计 & 优化 浏览次数:151次 腾讯大讲堂 2015年04月02日 字号: 大 中 小 分享到:QQ空间新浪微博腾讯微博人人网豆瓣网开心网更多0   编者按:经过2014年一年 ...

  8. LibreOJ #2006. 「SCOI2015」小凸玩矩阵

    想了挺久没想出来,一看题解恍然大悟.一个数对应一行和一列,二分答案,凡是小于等于答案的就连边.如果满足能够取出 \(n - k + 1\) 个不比二分中点 \(mid\) 大的数,那么r = mid, ...

  9. 代理实现aop以及代理工厂实现增强

    一.静态代理实现 1.接口(抽象主题) 2.接口的实现类(真实主题) 3.代理类(代理主题) 4.测试类: ApplicationContext context=new ClassPathXmlApp ...

  10. 【转载】将Centos的yum源更换为国内的阿里云源

    自己的yum源不知道什么时候给改毁了--搜到了个超简单的方法将yum源更换为阿里的源 完全参考 http://mirrors.aliyun.com/help/centos?spm=5176.bbsr1 ...