(CodeForces 548B 暴力) Mike and Fun
http://codeforces.com/problemset/problem/548/B
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.
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.
After each round, print the current score of the bears.
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
题意理解错, 唉, 白白错了两次, 粘个代码警示一下自己
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std; #define N 550 int v[N][N]; int Find(int a[], int m)
{
int i, num=, Max=;
for(i=; i<=m; i++)
{
if(a[i]==)
num++;
else
{
Max = max(Max, num);
num = ;
}
} Max = max(Max, num); return Max;
} int FindMax(int a[], int n)
{
int i, Max = ; for(i=; i<=n; i++)
Max = max(Max, a[i]); return Max;
} int main()
{
int n, m, q; while(scanf("%d%d%d", &n, &m, &q)!=EOF)
{
int i, j, I, J, a[N], Max; memset(a, , sizeof(a));
memset(v, , sizeof(v)); for(i=; i<=n; i++)
{
for(j=; j<=m; j++)
{
scanf("%d", &v[i][j]);
a[i] = Find(v[i], m);
}
} for(i=; i<=q; i++)
{
scanf("%d%d", &I, &J);
if(v[I][J])
{
v[I][J] = ;
a[I] = Find(v[I], m); Max = FindMax(a, n);
printf("%d\n", Max);
}
else
{
v[I][J] = ;
a[I] = Find(v[I], m); Max = FindMax(a, n);
printf("%d\n", Max);
}
}
}
return ;
}
(CodeForces 548B 暴力) Mike and Fun的更多相关文章
- Codeforces 547C/548E - Mike and Foam 题解
目录 Codeforces 547C/548E - Mike and Foam 题解 前置芝士 - 容斥原理 题意 想法(口胡) 做法 程序 感谢 Codeforces 547C/548E - Mik ...
- Codeforces 548B Mike and Fun
传送门 B. Mike and Fun time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- CodeForces 548B Mike and Fun (模拟)
题意:给定一个n*m的矩阵,都是01矩阵,然后每次一个询问,改变一个格的值,然后问你最大有数是多少. 析:就是按他说的模拟,要预处理,只要把每行的最大值记下来,当改变时,再更新这一行的最大值. 代码如 ...
- CodeForces 670D1 暴力或二分
今天,开博客,,,激动,第一次啊 嗯,,先来发水题纪念一下 D1. Magic Powder - 1 This problem is given in two versions that diff ...
- codeforces 361 A - Mike and Cellphone
原题: Description While swimming at the beach, Mike has accidentally dropped his cellphone into the wa ...
- Codeforces Round #439 (Div. 2) Problem E (Codeforces 869E) - 暴力 - 随机化 - 二维树状数组 - 差分
Adieu l'ami. Koyomi is helping Oshino, an acquaintance of his, to take care of an open space around ...
- Codeforces Round #439 (Div. 2) Problem A (Codeforces 869A) - 暴力
Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-sc ...
- Codeforces Round #425 (Div. 2) Problem B Petya and Exam (Codeforces 832B) - 暴力
It's hard times now. Today Petya needs to score 100 points on Informatics exam. The tasks seem easy ...
- Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem C (Codeforces 831C) - 暴力 - 二分法
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain ...
随机推荐
- IDEA配置maven,jdk,编码
1.电脑已解压maven,修改好settings.xml文件,本地仓库,下载远程镜像 2.
- 5D - Rectangles
Given two rectangles and the coordinates of two points on the diagonals of each rectangle,you have t ...
- JS-Date心得
写在前面的话:我觉得JavaScript以后也要出像Java1.8新增的Date类,这个月份要是从1开始多好,强迫症很难受.......也就发发牢骚,没出来之前还得这么用..... 属性: const ...
- 转录组分析---Hisat2+StringTie+Ballgown使用
转录组分析---Hisat2+StringTie+Ballgown使用 (2016-10-10 08:14:45) 转载▼ 标签: 生物信息学 转录组 1.Hisat2建立基因组索引: First ...
- NOIP需要掌握的内容(大致
1.排序算法(快排.选择.冒泡.堆排序.二叉排序树.桶排序)2.DFS/BFS 剪枝 哈希表3.树 ①遍历 ②二叉树 ③二叉排序树(查找.生成.删除) ④堆(二叉堆.左偏树.堆排序) ...
- linux内socket服务器无法连接windows
今天在试socket的时候出现了一个问题:问题概述是这样的: 1.linux采用centOS7(mini)版本,虚拟机版本VMware12,网卡设置NAT 2.服务器和客户端都在windows上,通讯 ...
- vue脚手架搭建的具体步骤
1.全局安装cli npm install -g vue-cli 在全局安装vue的命令行工具 2.初始化项目 vue init webpack my-project 初始化一个基于webpack ...
- 如何从jks文件中导出公私钥
1.从JKS转换到PKCS12 #keytool -importkeystore -srckeystore <MY_KEYSTORE.jks> -destkeystore <MY_F ...
- TensorFlow安装时错误CondaValueError: prefix already exists: G:\softs\Anaconda\envs\tensorflow
TensorFlow安装时,TensorFlow环境已经调好了,就是下面的第(3)步, 可我自己偏偏选了个Python3.7,因为检测到自己的Python最新版本为3.7,就手贱安了TensorFlo ...
- 【Linux】ApacheBench(ab)压力测试工具
AB的简介 ab是apachebench命令的缩写. ab是apache自带的压力测试工具.ab非常实用,它不仅可以对apache服务器进行网站访问压力测试,也可以对或其它类型的服务器进行压力测试.比 ...