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2017中国大学生程序设计竞赛 - 女生专场 1002 dp

2024-09-20 05:20:56   208人阅读

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 }

 

2017中国大学生程序设计竞赛 - 女生专场 1002 dp


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