Building Shops

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

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll inf=1ll<<60;
#define P pair<int,int>
const int N=3000+10;

P p[N];

ll dp[N][N];

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    //freopen("in.txt","r",stdin);
    //cout<<"1"<<endl;
    int n;
    set<P> s;
    while(cin>>n){
        for(int i=1;i<=n;i++){
            int x,y;
            cin>>x>>y;
            p[i]=P(x,y);
        }
        sort(p+1,p+1+n);
        dp[1][1]=p[1].second;
        for(int i=2;i<=n;i++){
            for(int j=1;j<i;j++)
                dp[i][j]=dp[i-1][j]+p[i].first-p[j].first;
            dp[i][i]=inf;
            for(int j=1;j<i;j++)
            dp[i][i]=min(dp[i][i],dp[i-1][j]+p[i].second);
        }
        ll ans=inf;
        for(int i=1;i<=n;i++)
        ans=min(ans,dp[n][i]);
        cout<<ans<<endl;
    }
}