Building Shops 
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) 
Total Submission(s): 701 Accepted Submission(s): 265

Problem Description 
HDU’s n classrooms are on a line ,which can be considered as a number line. Each classroom has a coordinate. Now Little Q wants to build several candy shops in these n classrooms.

The total cost consists of two parts. Building a candy shop at classroom i would have some cost ci. For every classroom P without any candy shop, then the distance between P and the rightmost classroom with a candy shop on P’s left side would be included in the cost too. Obviously, if there is a classroom without any candy shop, there must be a candy shop on its left side.

Now Little Q wants to know how to build the candy shops with the minimal cost. Please write a program to help him.

Input 
The input contains several test cases, no more than 10 test cases. 
In each test case, the first line contains an integer n(1≤n≤3000), denoting the number of the classrooms. 
In the following n lines, each line contains two integers xi,ci(−109≤xi,ci≤109), denoting the coordinate of the i-th classroom and the cost of building a candy shop in it. 
There are no two classrooms having same coordinate.

Output 
For each test case, print a single line containing an integer, denoting the minimal cost.

Sample Input 

1 2 
2 3 
3 4 

1 7 
3 1 
5 10 
6 1

Sample Output 

11

Source 
2017中国大学生程序设计竞赛 - 女生专场

题意: 
有n个教室,现在想在这n个教室中建一些超市,问你最少费用为多少? 
费用分为两种: 
1:在第i个教室建超市,费用因为ci 
2:没有建超市的教室的费用为它和它左边最接近的超市的坐标之间的距离

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
struct node
{
ll x;
ll c;
}a[];
bool cmp(node x,node y)
{
return x.x<y.x;
}
ll dp[];
//dp[i]表示只考虑前i个教室 的最优解
ll dp2[];
//dp2[i]表示 第i个固定条件下,前i个的最优解
ll s[];
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=;i<=n;i++)
{
scanf("%lld %lld",&a[i].x,&a[i].c); }
sort(a+,a+n+,cmp);
memset(dp,inf,sizeof(dp));
memset(dp2,inf,sizeof(dp2));
memset(s,,sizeof(s));
dp[]=;
dp[]=dp2[]=a[].c;
int pp=a[].x;
for(int i=;i<=n;i++)
{
a[i].x-=pp;
s[i]=a[i].x+s[i-];
}
for(int i=;i<=n;i++)
{
dp2[i]=dp[i-]+a[i].c;
//第i个固定条件下,前i个的最优解
for(int j=;j<=i;j++)
{
dp[i]=min(dp[i],dp2[j]+s[i]-s[j]-(i-j)*a[j].x);
}
}
printf("%lld\n",dp[n]);
}
return ;
}

Building Shops

Time
Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K
(Java/Others)
Total Submission(s): 2728    Accepted Submission(s):
936

Problem Description
HDU’s n

classrooms are on a line ,which can be considered as a number line. Each
classroom has a coordinate. Now Little Q wants to build several candy shops in
these n

classrooms.

The total cost consists of two parts. Building a candy shop
at classroom i

would have some cost ci

. For every classroom P

without any candy shop, then the distance between P

and the rightmost classroom with a candy shop on P

's left side would be included in the cost too. Obviously, if there is a
classroom without any candy shop, there must be a candy shop on its left
side.

Now Little Q wants to know how to build the candy shops with the
minimal cost. Please write a program to help him.

 
Input
The input contains several test cases, no more than 10
test cases.
In each test case, the first line contains an integer n(1≤n≤3000)

, denoting the number of the classrooms.
In the following n

lines, each line contains two integers xi,ci(−109≤xi,ci≤109)

, denoting the coordinate of the i

-th classroom and the cost of building a candy shop in it.
There are no two
classrooms having same coordinate.

 
Output
For each test case, print a single line containing an
integer, denoting the minimal cost.
 
Sample Input
3
1 2
2 3
3 4
4
1 7
3 1
5 10
6 1
 
Sample Output
5
11
 
Source
 
Recommend
jiangzijing2015   |   We have carefully selected
several similar problems for you:  6286 6285 6284 6283 6282 

2017中国大学生程序设计竞赛 - 女生专场(dp)的更多相关文章

  1. 2017中国大学生程序设计竞赛 - 女生专场 Deleting Edges(思维+最短路)

    Deleting Edges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  2. 2017中国大学生程序设计竞赛 - 女生专场 Happy Necklace(递推+矩阵快速幂)

    Happy Necklace Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  3. 2017中国大学生程序设计竞赛 - 女生专场(Graph Theory)

    Graph Theory Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)To ...

  4. 2017中国大学生程序设计竞赛 - 女生专场 1002 dp

    Building Shops Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  5. 2017中国大学生程序设计竞赛 - 女生专场B【DP】

    B HDU - 6024 [题意]:n个教室,选一些教室建造糖果商店. 每个教室,有一个坐标xi和在这个教室建造糖果商店的花费ci. 对于每一个教室,如果这个教室建造糖果商店,花费就是ci,否则就是与 ...

  6. 2017中国大学生程序设计竞赛 - 女生专场C【前后缀GCD】

    C HDU - 6025 [题意]:去除数列中的一个数字,使去除后的数列中所有数字的gcd尽可能大. [分析]: 数组prefixgcd[],对于prefixgcd[i]=g,g为a[0]-a[i]的 ...

  7. 2017中国大学生程序设计竞赛 - 女生专场A【模拟】

    A HDU - 6023 [题意]:求AC题数和总时长. [分析]:模拟.设置标记数组记录AC与否,再设置错题数组记录错的次数.罚时罚在该题上,该题没AC则不计入总时间,AC则计入.已经AC的题不用再 ...

  8. HDU 6024(中国大学生程序设计竞赛女生专场1002)

    这是CCPC女生专场的一道dp题.大佬们都说它简单,我并没有感到它有多简单. 先说一下题意:在一条直线上,有n个教室,现在我要在这些教室里从左到右地建设一些作为糖果屋,每个教室都有自己的坐标xi 和建 ...

  9. "巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

    Combine String #include<cstdio> #include<cstring> #include<iostream> #include<a ...

随机推荐

  1. Pandas描述性统计

    有很多方法用来集体计算DataFrame的描述性统计信息和其他相关操作. 其中大多数是sum(),mean()等聚合函数,但其中一些,如sumsum(),产生一个相同大小的对象. 一般来说,这些方法采 ...

  2. Eclipse_下载地址

    1. http://www.eclipse.org/downloads/ http://www.eclipse.org/downloads/packages/ http://archive.eclip ...

  3. Linux平台上DPDK入门指南

    1. 简介 本文档包含DPDK软件安装和配置的相关说明.旨在帮助用户快速启动和运行软件.文档主要描述了在Linux环境下编译和 运行DPDK应用程序,但是文档并不深入DPDK的具体实现细节. 1.1. ...

  4. LeetCode第[14]题(Java): Longest Common Prefix

    题目:最长公共前缀 难度:EASY 题目内容: Write a function to find the longest common prefix string amongst an array o ...

  5. uva 1511 最小生成树

    https://vjudge.net/problem/UVA-1151 题意,给出N个点以及二维坐标,可以在任意两点间建立通路,代价是两点欧几里得距离的平方,同时有q个套餐,套餐x有qx个点,代价是q ...

  6. python迭代器与生成器(二)

      一.什么是迭代? 迭代通俗的讲就是一个遍历重复的过程. 维基百科中 迭代(Iteration) 的一个通用概念是:重复某个过程的行为,这个过程中的每次重复称为一次迭代.具体对应到Python编程中 ...

  7. Unity3D事件顺序与功能

    Unity3D中所有控制脚本的基类MonoBehaviour有一些虚函数用于绘制中事件的回调,也可以直接理解为事件函数,例如大家都很清楚的Start,Update等函数,以下做个总结. Awake 当 ...

  8. windows下memcached安装以及php_memcache.dll扩展

    http://kimi.it/258.html http://kimi.it/259.html https://www.cnblogs.com/elenaPeng/p/6877530.html htt ...

  9. ElasticSearch安装及简单配置说明

      目录 1.      准备安装包... 1 2.      安装jdk7. 1 3.      安装ElasticSearch. 2 4.      安装maven. 3 5.      集成IK ...

  10. 【tensorflow:Google】二、Tensorflow环境搭建

    2.1 Tensorflow 主要依赖包 2.1.1 Protocol Buffer 结构化数据序列化的过程,另外的工具:XML, JSON, 区别:二进制(不可读):先定义数据格式,还原的时候将需要 ...