dp

HDU’s nn 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 nn classrooms. 

The total cost consists of two parts. Building a candy shop at classroom ii would have some cost cici. For every classroom PP without any candy shop, then the distance between PP and the rightmost classroom with a candy shop on PP'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. 

InputThe 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)n(1≤n≤3000), denoting the number of the classrooms. 
In the following nn lines, each line contains two integers xi,ci(−109≤xi,ci≤109)xi,ci(−109≤xi,ci≤109), denoting the coordinate of the ii-th classroom and the cost of building a candy shop in it. 
There are no two classrooms having same coordinate.OutputFor 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

      题意:给你n个教室，问在哪几个教室建立糖果屋可使得总花费最小，话费包括两部分1.在这个教室建立糖果屋花费ci，此教室不建糖果屋花费是这个教室左边到离他最近的糖果屋的距离。

每个教室只有建与不建两种情况，像背包。

假设左边最近超市为j，那么教室j+1~教室i都不能建超市，所以教室j+1~教室i的费用分别为他们的位置到教室j之间的距离了。当前dp[i][0] = dp[j][1] +( 教室j+1~教室i的费用)

如果我们暴力求解，那么时间复杂度会变成O(n^3)，会超时。但是我们会发现由于j是从大到小变化的，所以就可以用：t += (i - j) * (nodes[j+1].x - nodes[j].x);来记录教室j+1~教室i的费用和了。

关于 t += (i - j) * (nodes[j+1].x - nodes[j].x); 的解释： 
比如我们要算x3 - x1 ， x2 - x1的sum,那么由于保证了x是升序排列的，所以sum = (x3 - x2) + 2 * (x2 - x1).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

const int maxn = 3010;
const long long MM = 999999999999;
struct node
{
    long long x,c;
    bool operator < (node & a)
    {
        return this->x < a.x;
    }
    node():x(0),c(0){}
}nodes[maxn];

long long dp[maxn][2];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%I64d%I64d",&nodes[i].x,&nodes[i].c);
        }
        sort(nodes,nodes+n);
        dp[0][1] = nodes[0].c;
        dp[0][0] = MM;
        for(int i=1;i<n;i++)
        {
            dp[i][1] = nodes[i].c + min(dp[i-1][0],dp[i-1][1]);
            dp[i][0] = MM;
            long long t = 0;
            for(int j=i-1;j>=0;j--)
            {
                t += (i - j) * (nodes[j+1].x - nodes[j].x);
                dp[i][0] = min(dp[i][0],dp[j][1] + t);
            }
        }
        printf("%I64d\n",min(dp[n-1][0],dp[n-1][1]));
    }
    return 0;