2017中国大学生程序设计竞赛

Building Shops

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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 　　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

 

题意：给你n个教室的坐标和 当前教室建造糖果商店的代价  若当前教室不建造糖果商店则代价为 与左边最近的糖果商店的距离 第一个位置上的教室必然建造糖果商店 问最少的代价。

题解：dp[i][1]表示第i个教室建造糖果商店 前i个教室的最小代价,dp[i][2]表示第i个教室不建造糖果商店 前i个教室的最小代价,具体看代码。HDU 注意while多组输入 orz。

　

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
struct node
{
    ll x;
    ll c;
} N[3005];
bool cmp(struct node a,struct node b)
{
    return a.x<b.x;
}
ll dp[3005][5];
ll sum[3005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%I64d %I64d",&N[i].x,&N[i].c);
        sort(N+1,N+1+n,cmp);
        for(int i=1; i<=n; i++)
        {
            dp[i][1]=5e18;
            dp[i][2]=5e18;
        }
        dp[0][1]=0;
        dp[0][2]=0;
        for(int i=1; i<=n; i++)
        {
            dp[i][1]=min(dp[i-1][1],dp[i-1][2])+N[i].c;
            ll exm=0;
            for(int j=i-1; j>=1; j--)
            {
                exm=exm+(i-j)*(N[j+1].x-N[j].x);//累加距离
                dp[i][2]=min(dp[i][2],dp[j][1]+exm);
            }
        }
        printf("%I64d
",min(dp[n][1],dp[n][2]));
    }
    return 0;
}