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 }