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
2017中国大学生程序设计竞赛 - 女生专场

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
#include <vector>
#include<iostream>
using namespace std;
#define N 50005
#define INF 0x3f3f3f3f
#define LL long long
LL dp[N][2];
struct node
{
    LL x,c;
}s[N];
int cmp(node a,node b)
{
    return a.x<b.x;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%lld %lld",&s[i].x,&s[i].c);
        sort(s+1,s+n+1,cmp);
        memset(dp,0,sizeof(dp));
        dp[1][1] = s[1].c;///东边肯定有糖果店    所以第一个教室是必须建糖果店的
        dp[1][0] = INF;
        for(int i=2;i<=n;i++)
        {
            dp[i][1] = min(dp[i-1][1],dp[i-1][0])+s[i].c;
            ///在这个教室建立糖果店所需要的最小花费
            LL sum=0;
            dp[i][0] = INF;
            ///找到前面花费最小的糖果店
            for(int j=i-1;j>=1;j--)
            {
                sum+=(i-j)*(s[j+1].x-s[j].x);
                dp[i][0] = min(dp[i][0],dp[j][1]+sum);
            }
        }
        printf("%lld\n",min(dp[n][1],dp[n][0]));
        ///取建糖果店与不建糖果店的其中的小值
    }
    return 0;
}