Building Shops

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

Total Submission(s): 379    Accepted Submission(s): 144

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

// 一条直线上有 n 的教室，想要在这些点上建一些糖果店，建设糖果店的成本分为 2 部分，建设费，右边的非糖果店到这个糖果店的距离差的和(累加到是一个糖果店为止)

//典型DP题

dp[i] 为在 i 建造最后一个糖果店的最小花费的话

丛左到右 dp[i] = min(dp[i],dp[j]+shop[i].v-(n-i+1)*(shop[i].p-shop[j].p)) (1<=j<i) p是位置，v为建造费

还有就是需要排序，还有需要 long long 型

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
#define LL long long
#define MX 3005
struct Shop
{
    LL p,v;
    bool operator < (const Shop& b)const
    {
        return p<b.p;
    }
}shop[MX];
LL dp[MX];

int main()
{
    int n;
    while (scanf("%d",&n)!=EOF)
    {
        for (int i=1;i<=n;i++)
            scanf("%I64d%I64d",&shop[i].p,&shop[i].v);
        sort(shop+1,shop+1+n);
        LL total = 0;
        for (int i=1;i<=n;i++)
            total += shop[i].p - shop[1].p;

        dp[1]=shop[1].v+total;
        for (int i=2;i<=n;i++)
        {
            for (int j=1;j<i;j++)
            {
                if (j==1) dp[i] = dp[j] + shop[i].v - (n-i+1)*(shop[i].p-shop[j].p);
                else dp[i] = min(dp[i],dp[j]+shop[i].v-(n-i+1)*(shop[i].p-shop[j].p));
            }
        }
        LL ans = dp[1];
        for (int i=2;i<=n;i++)
            ans = min(dp[i],ans);
        printf("%I64d
",ans);
    }
    return 0;
}