[DP]Building Shops

Building Shops

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

正确解法：

有几个线性排列的空教室，现在准备把几个教室建造成糖果屋

1.每个教室建造糖果屋的成本为ci

2.空教室的成本是 与左边最近糖果屋的距离

我们用DP来做，f[i][1]表示第i个建造糖果屋，前i个的成本

f[i][0]表示第 i 个不建造糖果屋，前i个的成本。

很容易得出 f[i][1]=min(f[i-1][1],f[i-1][0])+ci；

f[i][0]=min(f[j][1]+  (j+1,j+2  ……,i 到j的距离）)

ll t=0;
for(int j=i-1;j>=1;j--)
{
    t+=(i-j)*(a[j+1].x-a[j].x);
    f[i][0]=min(f[i][0],f[j][1]+t);
}

最后，它的xi和ci 是（1，10^9） 开了 ll

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <cctype>
#include <sstream>
using namespace std;
typedef long long ll;
const int inf=0x7fffffff;
const int N=3000+100;
const int M=50000+10;
const int MOD=1e9+7;
const double PI=acos(-1.0);
int n;
ll f[N][3];
ll MM=999999999999;
struct node
{
    ll x,c;
}a[N];
bool cmp(node a,node b)
{
    return (a.x<b.x);
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        //f[0][1]=f[0][0]=0;
        for(int i=1;i<=n;i++)
            f[i][1]=f[i][0]=MM;
        for(int i=1;i<=n;i++)
            scanf("%lld %lld",&a[i].x,&a[i].c);
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
            f[i][1]=min(f[i-1][0],f[i-1][1])+a[i].c;
            ll t=0;
            for(int j=i-1;j>=1;j--)
            {
                t+=(i-j)*(a[j+1].x-a[j].x);
                f[i][0]=min(f[i][0],f[j][1]+t);
            }
        }
        printf("%lld
",min(f[n][1],f[n][0]));
    }


    return 0;
}