2017中国大学生程序设计竞赛 - 女生专场 1002 dp

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[];
 bool cmp(struct node a,struct node b)
 {
     return a.x<b.x;
 }
 ll dp[][];
 ll sum[];
 int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF)
     {
         for(int i=; i<=n; i++)
             scanf("%I64d %I64d",&N[i].x,&N[i].c);
         sort(N+,N++n,cmp);
         for(int i=; i<=n; i++)
         {
             dp[i][]=5e18;
             dp[i][]=5e18;
         }
         dp[][]=;
         dp[][]=;
         for(int i=; i<=n; i++)
         {
             dp[i][]=min(dp[i-][],dp[i-][])+N[i].c;
             ll exm=;
             for(int j=i-; j>=; j--)
             {
                 exm=exm+(i-j)*(N[j+].x-N[j].x);//累加距离
                 dp[i][]=min(dp[i][],dp[j][]+exm);
             }
         }
         printf("%I64d\n",min(dp[n][],dp[n][]));
     }
     return ;
 }