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=;i<=n;i++)
             scanf("%I64d%I64d",&shop[i].p,&shop[i].v);
         sort(shop+,shop++n);
         LL total = ;
         for (int i=;i<=n;i++)
             total += shop[i].p - shop[].p;

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