2017中国大学生程序设计竞赛 - 女生专场 1002 dp

Posted 2020-09-17 半根毛线code

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。

　

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define esp 0.00000000001
 5 struct node
 6 {
 7     ll x;
 8     ll c;
 9 } N[3005];
10 bool cmp(struct node a,struct node b)
11 {
12     return a.x<b.x;
13 }
14 ll dp[3005][5];
15 ll sum[3005];
16 int main()
17 {
18     int n;
19     while(scanf("%d",&n)!=EOF)
20     {
21         for(int i=1; i<=n; i++)
22             scanf("%I64d %I64d",&N[i].x,&N[i].c);
23         sort(N+1,N+1+n,cmp);
24         for(int i=1; i<=n; i++)
25         {
26             dp[i][1]=5e18;
27             dp[i][2]=5e18;
28         }
29         dp[0][1]=0;
30         dp[0][2]=0;
31         for(int i=1; i<=n; i++)
32         {
33             dp[i][1]=min(dp[i-1][1],dp[i-1][2])+N[i].c;
34             ll exm=0;
35             for(int j=i-1; j>=1; j--)
36             {
37                 exm=exm+(i-j)*(N[j+1].x-N[j].x);//累加距离
38                 dp[i][2]=min(dp[i][2],dp[j][1]+exm);
39             }
40         }
41         printf("%I64d\n",min(dp[n][1],dp[n][2]));
42     }
43     return 0;
44 }