Art Union
A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.
Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.
Order is important everywhere, so the painters' work is ordered by the following rules:
- Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);
- each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;
- each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;
- as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.
Given that the painters start working at time 0, find for each picture the time when it is ready for sale.
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.
5 1
1
2
3
4
5
1 3 6 10 15
4 2
2 5
3 1
5 3
10 1
7 8 13 21
做题像是憋屎,想明白后,一下就出来了。
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <unordered_set>
#include <unordered_map>
#include <xfunctional>
#define ll long long
#define mod 998244353
using namespace std;
int dir[][] = { {,},{,-},{-,},{,} };
const int maxn = ;
const long long inf = 0x7f7f7f7f7f7f7f7f; int main()
{
int m, n;
cin >> m >> n;
vector<vector<int>> t(m+, vector<int>(n+,));
for (int i = ; i <= m; i++)
for (int j = ; j <= n; j++)
cin >> t[i][j];
for (int i = ; i <= m; i++)
{
for (int j = ; j <= n; j++)
{
if (t[i][j - ] < t[i - ][j])
t[i][j] += t[i - ][j];
else
t[i][j] += t[i][j - ];
}
}
for (int i = ; i <= m; i++)
{
cout << t[i][n] << " ";
}
return ;
}
Art Union的更多相关文章
- Codeforces Round #241 (Div. 2)->B. Art Union
B. Art Union time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- Codeforces Round #241 (Div. 2) B. Art Union 基础dp
B. Art Union time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- Codeforces Round #241 (Div. 2) B. Art Union (DP)
题意:有\(n\)个画家,\(m\)幅画,每个画家负责\(m\)幅画,只有前一个画家画完时,后面一个画家才能接着画,一个画家画完某幅画的任务后,可以开始画下一幅画的任务,问每幅画最后一个任务完成时的时 ...
- n个人作m幅画
题目网址: http://codeforces.com/problemset/problem/416/B A well-known art union called "Kalevich is ...
- Codeforces Round #241 (Div. 2) B dp
B. Art Union time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- The Art of Deception
前言 一些黑客毁坏别人的文件甚至整个硬盘,他们被称为电脑狂人(crackers)或计算机破坏者(vandals).另一些新手省去学习技术的麻烦,直接下载黑客工具侵入别人的计算机,这些人被称为脚本小子( ...
- Android内存优化(一)DVM和ART原理初探
相关文章 Android内存优化系列 Java虚拟机系列 前言 要学习Android的内存优化,首先要了解Java虚拟机,此前我用了多篇文章来介绍Java虚拟机的知识,就是为了这个系列做铺垫.在And ...
- SQL Server-聚焦UNIOL ALL/UNION查询(二十三)
前言 本节我们来看看有关查询中UNION和UNION ALL的问题,简短的内容,深入的理解,Always to review the basics. 初探UNION和UNION ALL 首先我们过一遍 ...
- SQL 提示介绍 hash/merge/concat union
查询提示一直是个很有争议的东西,因为他影响了sql server 自己选择执行计划.很多人在问是否应该使用查询提示的时候一般会被告知慎用或不要使用...但是个人认为善用提示在不修改语句的条件下,是常用 ...
随机推荐
- NIO学习笔记,从Linux IO演化模型到Netty—— Linux零拷贝
这里只是感性地认识Linux零拷贝,不涉及具体细节. 1.Linux传统的数据拷贝 用户进程是不能直接访问文件系统的,要先切换到内核态,发起系统调用,DMA把磁盘中的数据写入内核空间,内核再把数据拷贝 ...
- Linux内核本地提权漏洞(CVE-2019-13272)
漏洞描述 kernel / ptrace.c中的ptrace_link错误地处理了想要创建ptrace关系的进程的凭据记录,这允许本地用户通过利用父子的某些方案来获取root访问权限 进程关系,父进程 ...
- 如何在vue-cli中使用vuex(配置成功
前言 众所周知,vuex 是一个专为 vue.js 应用程序开发的状态管理模式,在构建一个中大型单页应用中使用vuex可以帮助我们更好地在组件外部管理状态.而vue-cli是vue的官方脚手架,它能帮 ...
- 如何将旧Mac的数据迁移到新的MacBook Pro?
最新版的MacBook Pro已经上市,具有超凡魅力的Touch Bar开创了一个新时代.苗条的设计和华丽的显示效果也起到了推动运动的作用……!将数据从旧Mac传输到新Mac不再是一件漫长的事.您只需 ...
- fatal error LNK1169: one or more multiply defined symbols found
在 Project/Setting/Link/General中的 Project Options: 加入 /FORCE:MULTIPLE即可")可以解决报错问题,但是这些问题全部变成了war ...
- Wannafly Winter Camp 2020 Day 5C Self-Adjusting Segment Tree - 区间dp,线段树
给定 \(m\) 个询问,每个询问是一个区间 \([l,r]\),你需要通过自由地设定每个节点的 \(mid\),设计一种"自适应线段树",使得在这个线段树上跑这 \(m\) 个区 ...
- ActiveMQ的JMS消息可靠机制
JMS消息可靠机制 ActiveMQ消息签收机制: 客戶端成功接收一条消息的标志是一条消息被签收,成功应答. 消息的签收情形分两种: 1.带事务的session 如果session带有事务,并且事务成 ...
- uniGUI之自定义JS事件动作ClientEvents(30)
sender.setText('Over'); MainForm.UniLabel1.setPosition(x, y); MainForm.UniLabel1.setText('Click!');
- react-发表评论案例
评论列表组件 import React from 'react' import CMTItem from './CmtItem.jsx' import CMTBox from './CmtBox.js ...
- archlinux下安装nvidia驱动解决Nvidia显卡显示问题
root下使用以下命令: sudo pacman -S nvidia nvidia-libgl