Problem H: Heaps

Time Limit: 2 Sec  Memory Limit: 128 MB
Submit: 48  Solved: 9
[Submit][Status][Web Board]

Description

Zuosige always has bad luck. Recently, he is in hospital because of pneumonia. While he is taking his injection, he feels extremely bored. However, clever Zuosige comes up with a new game.

Zuosige knows there is a typical problem called Merging Stones. In the problem, you have N heaps of stones and you are going to merging them into one heap. The only restriction is that you can only merging adjacent heaps and the cost of a merging operation is the total number of stones in the two heaps merged. Finally, you are asked to answer the minimum cost to accomplish the merging.

However, Zuosige think this problem is too simple, so he changes it. In his problem, the cost of a merging is a polynomial function of the total number of stones in those two heaps and you are asked to answer the minimum cost.

Input

The first line contains one integer T, indicating the number of test cases.
In one test case, there are several lines.
In the first line, there are an integer N (1<=N<=1000).
In the second line, there are N integers. The i-th integer si (1<=si<=40) indicating the number of stones in the i-th heap.
In the third line, there are an integer m (1<=m<=4).
In the forth line, there are m+1 integers a0, … , am. The polynomial function is P(x)= (a0+a1*x+a2*x2+…+am*xm). (1<=ai<=5)

Output

For each test case, output an integer indicating the answer.

Sample Input

1
5
3 1 8 9 9
2
2 1 2

Sample Output

2840

HINT

转载请注明出处:http://www.cnblogs.com/yuyixingkong/

题目链接:http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1616

 #include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define N 1005
#define mod 19999997
#define INF 0x3f3f3f3f
#define exp 1e-8 LL dp[N][N],vec[],sum[N];
int s[N],a[N],t,n,m,tot,vis[N][N]; LL col(LL x)
{
LL ans = a[];
int i,j;
up(i,,m)
{
LL tem = ;
up(j,,i) tem*=x;
ans+=tem*a[i];
}
return ans;
} int main()
{
int i,j,k;
scanf("%d",&t);
w(t--)
{
scanf("%d",&n);
mem(sum,);
mem(dp,);
tot=;
up(i,,n)
{
scanf("%d",&s[i]);
sum[i] = sum[i-]+s[i];
tot+=s[i];
}
scanf("%d",&m);
up(i,,m)
{
scanf("%d",&a[i]);
}
up(i,,tot)
{
vec[i]=col((LL)i);
}
up(i,,n) vis[i][i] = i;
vis[][] = ;
int len;
up(len,,n)
{
up(i,,n-len+)
{
j = i+len-;
dp[i][j] = 1LL<<;
up(k,vis[i][j-],vis[i+][j])
{
LL tem = dp[i][k]+dp[k+][j]+vec[sum[j]-sum[i-]];
if(tem<dp[i][j])
{
dp[i][j] = tem;
vis[i][j] = k;
}
}
}
}
printf("%lld\n",dp[][n]);
} return ;
} /**************************************************************
Problem: 1616
User: aking2015
Language: C++
Result: Accepted
Time:620 ms
Memory:13728 kb
****************************************************************/

Heaps(Contest2080 - 湖南多校对抗赛(2015.05.10)(国防科大学校赛决赛-Semilive)+scu1616)的更多相关文章

  1. Contest2073 - 湖南多校对抗赛(2015.04.06)

    Contest2073 - 湖南多校对抗赛(2015.04.06) Problem A: (More) Multiplication Time Limit: 1 Sec  Memory Limit:  ...

  2. Contest2071 - 湖南多校对抗赛(2015.03.28)

    Contest2071 - 湖南多校对抗赛(2015.03.28) 本次比赛试题由湖南大学ACM校队原创 http://acm.csu.edu.cn/OnlineJudge/contest.php?c ...

  3. 湖南多校对抗赛(2015.05.03)Problem A: Twenty-four point

    给四个数 问能不能算出24点...我的方法比较烂...920ms 差点TLE.应该有更好的方法. #include<stdio.h> #include<string.h> #i ...

  4. 湖南多校对抗赛(2015.05.03)Problem B: War

    并查集.从后往前加边. #include<stdio.h> #include<string.h> #include<math.h> #include<algo ...

  5. Contest2089 - 湖南多校对抗赛(2015.05.31) Swipe(csu1648)

    Problem E: Swipe Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 100  Solved: 15[Submit][Status][Web ...

  6. Aquarium Tank(csu1634+几何+二分)Contest2087 - 湖南多校对抗赛(2015.05.24)-G

    Aquarium Tank Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 15  Solved: 4[Submit][Status][Web Board ...

  7. Clock Pictures(kmp + Contest2075 - 湖南多校对抗赛(2015.04.26))

    Problem H: Clock Pictures Time Limit: 3 Sec  Memory Limit: 128 MBSubmit: 73  Solved: 18[Submit][Stat ...

  8. CSU 2136 ——湖南多校对抗赛 I

    2136: 统帅三军! Submit Page   Summary   Time Limit: 1 Sec     Memory Limit: 128 Mb     Submitted: 55     ...

  9. 2015.05.15,外语,学习笔记-《Word Power Made Easy》 01 “如何讨论人格特点”

    2015.03.17,外语,读书笔记-<Word Power Made Easy> 01 “如何讨论人格特点”学习笔记 SESSIONS 1 本来这些章节都是在一两年前学习的,现在趁给友人 ...

随机推荐

  1. [转]Linux操作系统tcpdump抓包分析详解

    PS:tcpdump是一个用于截取网络分组,并输出分组内容的工具,简单说就是数据包抓包工具.tcpdump凭借强大的功能和灵活的截取策略,使其成为Linux系统下用于网络分析和问题排查的首选工具. t ...

  2. 【webpack】-- 入门与解析

    每次学新东西总感觉自己是不是变笨了,看了几个博客,试着试着就跑不下去,无奈只有去看官方文档. webpack是基于node的.先安装最新的node. 1.初始化 安装node后,新建一个目录,比如ht ...

  3. Flutter介绍 - Flutter,H5,React Native之间的对比

    Flutter介绍 Flutter是Google推出的开源移动应用开发框架.开发者可以通过开发一套代码同时运行在iOS和Android平台. 它使用Dart语言进行开发,并且最终编译成各个平台的Nat ...

  4. Javascript高级编程学习笔记(40)—— DOM(6)CDATASection、DocumentType

    CDATASection类型 CDATASection类型是只针对XML文档的类型 因为浏览器无法解析 在浏览器中创建CDATASection的函数也无法正常使用 该类型有以下属性 nodeType: ...

  5. Python的 is 和 == 弄懂了吗?

    在Python中一切都是对象. Python中对象包含的三个基本要素,分别是: id(身份标识) type(数据类型) value(值) 对象之间比较是否相等可以用 == ,也可以用 is . is ...

  6. NuGet 构建服务器与常用命令

    公司出于某些原因需要自己在内部网络搭建一个私有的 Nuget 服务器,而且要运行在 Linux服务器上面.如果说 Windows 下搭建的话很简单,直接在项目当中引入 Nuget 的库就 OK,这儿的 ...

  7. HTTP/2协议–特性扫盲篇

    HTTP/2协议–特性扫盲篇 随着web技术的飞速发展,1999年制定的HTTP 1.1已经无法满足大家对性能的要求,Google推出协议SPDY,旨在解决HTTP 1.1中广为人知的性能问题.SPD ...

  8. Vue : Expected the Promise rejection reason to be an Error

    在vue项目中添加ESLint,new 一个 Promise 一直显示错误 :Expected the Promise rejection reason to be an Error 正常来说new ...

  9. Kubernetes 服务入口管理与 Nginx Ingress Controller

    Kubernetes 具有强大的副本,动态扩容等特性,每一次 Pod 的变化 IP 地址都会发生变化,所以 Kubernetes 引进了 Service 的概念.Kubernetes 中使用 Serv ...

  10. cp2102 驱动 win7x64 -2018

    试了好多种网上的驱动,都不行,要么是报错要么是安装没反应 之后意外遇见驱动官网?里面真全 url:https://www.silabs.com/products/development-tools/s ...