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.
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.
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 ",min(dp[n][1],dp[n][2])); 42 } 43 return 0; 44 }