B. Mike and Fun

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/548/problem/A

Description

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 nm 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.

Sample 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

Sample Output

3
4
3
3
4
 #include<bits/stdc++.h>
using namespace std; int main() {
int n,m,q,x,y;
int mapp[][];
cin>>n>>m>>q;
for(int i = ; i <= n; i++) {
int total = ;
int maxx = ;
for(int j = ; j <= m; j++) {
scanf("%d",&mapp[i][j]);
if(mapp[i][j] == )
total = ;
total += mapp[i][j];
/**更新该行最大的连续值*/
if(total > maxx) maxx = total;
}
mapp[i][] = maxx;
} int total;
while(q--) {
scanf("%d %d",&x,&y);
/**依题意进行处理*/
if(mapp[x][y] == ) {
mapp[x][y] = ;
} else {
mapp[x][y] = ;
} int maxx = ;
total = ; /**更新该行最大的连续值*/
for(int i = ; i <= m; i++) {
total += mapp[x][i];
if(total > maxx) maxx = total;
if(mapp[x][i] == ) total = ;
} /**找最大的连续值*/
mapp[x][] = maxx;
int ans = mapp[][];
for(int i = ; i <= n; i++) {
if(mapp[i][] > ans) ans = mapp[i][];
}
cout<<ans<<endl;
} return ;
}

Codeforces Round #305 (Div. 2) B. Mike and Fun 暴力的更多相关文章

  1. Codeforces Round #305 (Div. 2) A. Mike and Fax 暴力回文串

     A. Mike and Fax Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/548/pro ...

  2. Codeforces Round #305 (Div. 1) A. Mike and Frog 暴力

     A. Mike and Frog Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/547/pr ...

  3. set+线段树 Codeforces Round #305 (Div. 2) D. Mike and Feet

    题目传送门 /* 题意:对于长度为x的子序列,每个序列存放为最小值,输出长度为x的子序列的最大值 set+线段树:线段树每个结点存放长度为rt的最大值,更新:先升序排序,逐个添加到set中 查找左右相 ...

  4. 数论/暴力 Codeforces Round #305 (Div. 2) C. Mike and Frog

    题目传送门 /* 数论/暴力:找出第一次到a1,a2的次数,再找到完整周期p1,p2,然后以2*m为范围 t1,t2为各自起点开始“赛跑”,谁落后谁加一个周期,等到t1 == t2结束 详细解释:ht ...

  5. 暴力 Codeforces Round #305 (Div. 2) B. Mike and Fun

    题目传送门 /* 暴力:每次更新该行的num[],然后暴力找出最优解就可以了:) */ #include <cstdio> #include <cstring> #includ ...

  6. 字符串处理 Codeforces Round #305 (Div. 2) A. Mike and Fax

    题目传送门 /* 字符串处理:回文串是串联的,一个一个判断 */ #include <cstdio> #include <cstring> #include <iostr ...

  7. Codeforces Round #305 (Div. 1) B. Mike and Feet 单调栈

    B. Mike and Feet Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/547/pro ...

  8. Codeforces Round #305 (Div. 2) D. Mike and Feet 单调栈

    D. Mike and Feet time limit per test 1 second memory limit per test 256 megabytes input standard inp ...

  9. Codeforces Round #305 (Div. 2) D. Mike and Feet

    D. Mike and Feet time limit per test 1 second memory limit per test 256 megabytes input standard inp ...

随机推荐

  1. POJ-1182 食物链---并查集(附模板)

    题目链接: https://vjudge.net/problem/POJ-1182 题目大意: 中文题,不多说. 思路: 给每个动物创建3个元素,i-A, i-B, i-C i-x表示i属于种类x,并 ...

  2. Linux服务器SSH无法通过DSA证书登录的解决方法

    从openssh7.0开始,ssh-dss密钥被默认禁用. 修改服务器端的openssh设置重新开启 # vim /etc/sshd/sshd_config添加以下选项PubkeyAcceptedKe ...

  3. Dev GridControl GridView 属性大全[中文解释]

    Options 选项 OptionsBehavior 视图的行为选项 AllowAddRows 允许添加新数据行 AllowDeleteRows 允许删除数据行 AllowIncrementalSea ...

  4. 【MySQL】通过Binary Log简单实现数据回滚(一)

    一.前言 对,没错,我又水了好一阵子,深刻反思寄几.前段时间,工作项目上出于对excel等批量操作可能出现误操作的问题,要求提供一个能够根据操作批次进行数据回滚的能力.在开发的过程中接触到了MySQL ...

  5. C#之冒泡排序

    以前在学校的时候看过冒泡排序,看的时候挺明白的,但是自己写的时候就写不出来,在网上搜索了一下,发现网上的冒泡排序算法几乎都不符合冒泡排序的原理,虽然也能实现,但是不正宗. 冒泡排序从字面意思理解:应该 ...

  6. OpenGL平面阴影

    几种绘制阴影的方法 在OpenGL中,比较常见的绘制阴影的方法有:shadow mapping,shadow volumes以及一种在红宝书上提及的适合在确定平面上绘制阴影的方法. 平面阴影 在确定的 ...

  7. 《c++ const 详细总结》--转载

    C++中的const关键字的用法非常灵活,而使用const将大大改善程序的健壮性,本人根据各方面查到的资料进行总结如下,期望对朋友们有所帮助. const 是C++中常用的类型修饰符,常类型是指使用类 ...

  8. STL deque

      STL之deque容器详解 Deque 容器 deque容器是C++标准模版库(STL,Standard Template Library)中的部分内容.deque容器类与vector类似,支持随 ...

  9. [HNOI 2017]抛硬币

    Description 题库链接 两人抛硬币一人 \(a\) 次,一人 \(b\) 次.记正面朝上多的为胜.问抛出 \(a\) 次的人胜出的方案数. \(1\le a,b\le 10^{15},b\l ...

  10. UVA 3485 Bridge

    题目大意 你的任务是修建一座大桥.桥上等距地摆放着若干个塔,塔高为H,宽度忽略不计.相邻两座塔之间的距离不能超过D.塔之间的绳索形成全等的对称抛物线.桥长度为B,绳索总长为L,如下图所示求建最少的塔时 ...