Codeforces Round #305 (Div. 2) B. Mike and Fun 暴力
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 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.
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 使用 RHEL(RedHat)6.1 iso 安装包 安装Samba过程
今天因为工作的需要安装了(RHEL)redhat 6.1 自己为了方便就安装Samba 以记之. 注:Linux系统是刚刚安装好的所以没有samba安装的任何记录. 安装准备: ISO:RHEL_6. ...
- requests-模拟登陆
import requests requests.get('http://httpbin.org/cookies/set/number/123456') response = requests.get ...
- EF5中 执行 sql语句使用Database.ExecuteSqlCommand 返回影响的行数 ; EF5执行sql查询语句 Database.SqlQuery 带返回值
一: 执行sql语句,返回受影响的行数 在mysql里面,如果没有影响,那么返回行数为 -1 ,sqlserver 里面 还没有测试过 using (var ctx = new MyDbConte ...
- [Nginx]-外部多端口映射Https443端口配置
https服务器配置完成后,域名访问默认匹配至443端口,如果想同时通过https域名网址来请求多个对外服务,就需要在Nginx配置里来对请求进行规则判断,并匹配至相应的内部端口,这也是Nginx反向 ...
- 理解error和exception之间的区别
很多程序员不清楚error和exception之间的区别,这区别对于如何正确的处理问题而言非常重要(见附1,"简要的叙述error和exception").就像Mary Campi ...
- [LeetCode] Number Of Corner Rectangles 边角矩形的数量
Given a grid where each entry is only 0 or 1, find the number of corner rectangles. A corner rectang ...
- dropzone.js使用实践
官网地址:http://www.dropzonejs.com/ 一,它是什么: DropzoneJS is an open source library that provides drag'n'dr ...
- wows
[问题描述]山山最近在玩一款游戏叫战舰世界(steam 游戏太少了),他被大舰巨炮的魅力折服,于是山山开了一局游戏,这次发现目标是一艘战列舰新墨西哥级,舰桥很高,原本应该打在目标身后的圆形水域内的炮弹 ...
- [HNOI2011]赛车游戏
题目描述 名歌手LAALA最近迷上了一款赛车游戏,游戏中开车的玩家在不同的路段需要选择不同的速度,使得自己在最短的时间内到达终点.开始游戏时,车内的初始油量为f,所以游戏的关键是如何在速度和耗油量之间 ...
- TopCoder SRM 560 Div 1 - Problem 1000 BoundedOptimization & Codeforces 839 E
传送门:https://284914869.github.io/AEoj/560.html 题目简述: 定义"项"为两个不同变量相乘. 求一个由多个不同"项"相 ...