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

女生赛的题目也做不出来….
这题题解都是用dp dp[i][0/1]表示第i个取或不取的最小花费

我本来用的贪心 没过
注意代码里那个距离 并不是单纯指两个地点之间的距离 他是有一个累加的 原因就在注释里 因为当这个点i不建 而建的是前面的一个点j 那么j和i之间有可能还有房子没建（必须没建） 所以还要累加上这些点到j的距离

代码;

#include <cstdio>
#include <algorithm>
#include <cstring>
typedef long long ll;
using namespace std;
const int N=3005;
const int INF=0x3f3f3f3f;
ll dp[N][2];
ll dis[N][N];
struct node{
    int a,c;
}b[N];
bool cmp(node a,node b){
    return a.a<b.a;
}
int main(void){
    int n;
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++){
            scanf("%d%d",&b[i].a,&b[i].c);
        }
        sort(b,b+n,cmp);
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                dis[j][i]=dis[j-1][i]+b[j].a-b[i].a;
            }
        }
        for(int i=0;i<n;i++){
            dp[i][0]=dp[i][1]=INF;
        }
        dp[0][1]=b[0].c;
        ll ans=INF;
        for(int i=1;i<n;i++){
            dp[i][1]=min(dp[i-1][0],dp[i-1][1])+b[i].c;
            for(int j=i-1;j>=0;j--){
                dp[i][0]=min(dp[j][1]+dis[i][j],dp[i][0]);
            }
            //printf("%d %d %d\n",i,dp[i][0],dp[i][1]);
        }
        ans=min(dp[n-1][0],dp[n-1][1]);
        printf("%lld\n",ans);
    }
    return 0;
}
// #include <cstdio>
// #include <algorithm>
// using namespace std;
// const int N=3005;
// int a[N];
// int c[N];
// int r[N];
// int n;
// bool cmp(int i,int j){
   
// return a[i]<a[j];
// }
// int main(void){
   
// while(~scanf("%d",&n)){
   
// for(int i=0;i<n;i++){
   
// r[i]=i;
// scanf("%d%d",&a[i],&c[i]);
// }
// sort(r,r+n,cmp);
// int p=a[0];
// long long ans=c[0];
// for(int i=1;i<n;i++){
   
// if(a[r[i]]-p<c[r[i]]){
   
// ans+=(a[r[i]]-p);
// }
// else{
   
// p=a[r[i]];
// ans+=c[r[i]];
// }
// }
// printf("%lld\n",ans);
// }
// return 0;
// }