codeforces548B
Mike and Fun
Mike and some bears are playing a game just for fun. Mike is the judge. All bears except Mike are standing in an n × m grid, there's exactly one bear in each cell. We denote the bear standing in column number j of row number i by (i, j). Mike's hands are on his ears (since he's the judge) and each bear standing in the grid has hands either on his mouth or his eyes.
They play for q rounds. In each round, Mike chooses a bear (i, j) and tells him to change his state i. e. if his hands are on his mouth, then he'll put his hands on his eyes or he'll put his hands on his mouth otherwise. After that, Mike wants to know the score of the bears.
Score of the bears is the maximum over all rows of number of consecutive bears with hands on their eyes in that row.
Since bears are lazy, Mike asked you for help. For each round, tell him the score of these bears after changing the state of a bear selected in that round.
Input
The first line of input contains three integers n, m and q (1 ≤ n, m ≤ 500 and 1 ≤ q ≤ 5000).
The next n lines contain the grid description. There are m integers separated by spaces in each line. Each of these numbers is either 0 (for mouth) or 1 (for eyes).
The next q lines contain the information about the rounds. Each of them contains two integers i and j (1 ≤ i ≤ n and 1 ≤ j ≤ m), the row number and the column number of the bear changing his state.
Output
After each round, print the current score of the bears.
Examples
5 4 5
0 1 1 0
1 0 0 1
0 1 1 0
1 0 0 1
0 0 0 0
1 1
1 4
1 1
4 2
4 3
3
4
3
3
4 sol:直接暴力模拟就可以了,每次修改一个点,只会影响一行的答案,所以只要修改一行就可以了,O(Q*m)水过。。。
#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
ll s=;
bool f=;
char ch=' ';
while(!isdigit(ch))
{
f|=(ch=='-'); ch=getchar();
}
while(isdigit(ch))
{
s=(s<<)+(s<<)+(ch^); ch=getchar();
}
return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
if(x<)
{
putchar('-'); x=-x;
}
if(x<)
{
putchar(x+''); return;
}
write(x/);
putchar((x%)+'');
return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int N=;
int n,m,Q,a[N][N];
int ans[N];
int main()
{
int i,j;
R(n); R(m); R(Q);
for(i=;i<=n;i++)
{
for(j=;j<=m;j++) R(a[i][j]);
}
for(i=;i<=n;i++)
{
int tmp=;
for(j=;j<=m;j++)
{
if(a[i][j]) tmp++;
else tmp=;
ans[i]=max(ans[i],tmp);
}
}
while(Q--)
{
int x,y,Max=;
R(x); R(y);
a[x][y]^=;
int tmp=; ans[x]=;
for(i=;i<=m;i++)
{
if(a[x][i]) tmp++;
else tmp=;
ans[x]=max(ans[x],tmp);
}
for(i=;i<=n;i++) Max=max(Max,ans[i]);
Wl(Max);
}
return ;
}
/*
input
5 4 5
0 1 1 0
1 0 0 1
0 1 1 0
1 0 0 1
0 0 0 0
1 1
1 4
1 1
4 2
4 3
output
3
4
3
3
4
*/
codeforces548B的更多相关文章
随机推荐
- java 迭代器遍历List Set Map
Iterator接口: 所有实现了Collection接口的容器类都有一个iterator方法用以返回一个实现Iterator接口的对象 Iterator对象称作为迭代器,用以方便的对容器内元素的遍历 ...
- RBAC 基于权限的访问控制 serviceaccount -- clusterRole clusterRoleBinding
1.Role , RoleBinding 的作用对象都是namespace. 2.通过RoleRef,可以看到,RoleBinding对象通过名字,直接引用前面定义的Role,实现subject(us ...
- SQL Server-聚焦深入理解死锁以及避免死锁建议(转载)
前言 终于进入死锁系列,前面也提到过我一直对隔离级别和死锁以及如何避免死锁等问题模棱两可,所以才鼓起了重新学习SQL Server系列的勇气,本节我们来讲讲SQL Server中的死锁,看到许多文章都 ...
- Luogu4249 WC2007 石头剪刀布 费用流
传送门 考虑竞赛图三元环计数,设第\(i\)个点的入度为\(d_i\),根据容斥,答案为\(C_n^3 - \sum C_{d_i}^2\) 所以我们需要最小化\(\sum C_{d_i}^2\) 考 ...
- C# 双击ListView出现编辑框可编辑,回车确认
原文:C# 双击ListView出现编辑框可编辑,回车确认 //获取鼠标点击的项------API [DllImport("user32")] public static exte ...
- git 提交新增文件到网站
git add -A 是将所有的修改都提交.你可以用git status查看当前的变化,然后通过git add xxx有选择的提交.git commit 是将变化先提交到本地.git commit - ...
- 一次永久解决cmd窗口汉字显示乱码
对于编译出的程序,在 cmd 和 power shell 运行时都不能正确显示汉字. 网上查,可以再命令窗口修改: 1.打开CMD.exe命令行窗口 2.通过 chcp命令改变代码页,UTF-8的代码 ...
- 机器学习 第五篇:分类(kNN)
K最近邻(kNN,k-NearestNeighbor)算法是一种监督式的分类方法,但是,它并不存在单独的训练过程,在分类方法中属于惰性学习法,也就是说,当给定一个训练数据集时,惰性学习法简单地存储或稍 ...
- 2018年高教社杯全国大学生数学建模竞赛C题解题思路
题目 C题 大型百货商场会员画像描绘 在零售行业中,会员价值体现在持续不断地为零售运营商带来稳定的销售额和利润,同时也为零售运营商策略的制定提供数据支持.零售行业会采取各种不同方法来吸引更多的人成 ...
- java注解XML
用的是jdk自带的javax.xml.bind.JAXBContext将对象和xml字符串进行相互转换. 比较常用的几个: @XmlRootElement:根节点 @XmlAttribute:该属性作 ...