（hdu 6024） Building Shops

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6024

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  test cases.
In each test case, the first line contains an integer n(≤n≤), denoting the number of the classrooms.
In the following n lines, each line contains two integers xi,ci(−≤xi,ci≤), 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

Sample Output

Source
2017中国大学生程序设计竞赛 - 女生专场

题目大意：有N个教室，需要在教室建糖果店，所需要的花费有两种情况

1.建糖果店所需要花费的费用。

2.这个点离左边最近的糖果店的距离。

分析：这个点有两种可能  建糖果店  不建糖果店 所以是道dp题

设dp【i】【0】表示在这个点建立糖果店所需要花费的最小费用，dp【i】【0】在这个点不建糖果店所需要花费的最小费用

dp【i】【1】=min（dp[i-1][1]，dp[i-1][0]）+ci;

dp[i][0]的情况需要从起点到开始枚举，所以时间复杂度为O(n*n);

#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][];
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=;i<=n;i++)
            scanf("%lld %lld",&s[i].x,&s[i].c);
        sort(s+,s+n+,cmp);
        memset(dp,,sizeof(dp));
        dp[][] = s[].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[][] = INF;
        for(int i=;i<=n;i++)
        {
            dp[i][] = min(dp[i-][],dp[i-][])+s[i].c;
            ///在这个教室建立糖果店所需要的最小花费
            LL sum=;
            dp[i][] = INF;
            ///找到前面花费最小的糖果店
            for(int j=i-;j>=;j--)
            {
                sum+=(i-j)*(s[j+].x-s[j].x);
                dp[i][] = min(dp[i][],dp[j][]+sum);
            }
        }
        printf("%lld\n",min(dp[n][],dp[n][]));
        ///取建糖果店与不建糖果店的其中的小值
    }
    return ;
}