2017中国大学生程序设计竞赛 - 女生专场（dp）
Building Shops
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 701 Accepted Submission(s): 265
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 candy 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
Source
2017中国大学生程序设计竞赛 - 女生专场
题意：
有n个教室，现在想在这n个教室中建一些超市，问你最少费用为多少？
费用分为两种：
1：在第i个教室建超市，费用因为ci
2：没有建超市的教室的费用为它和它左边最接近的超市的坐标之间的距离
#include <iostream> #include<cstring> #include<string> #include<cstdio> #include<algorithm> #include<cmath> #include<deque> #define ll long long #define inf 0x3f3f3f3f using namespace std; struct node { ll x; ll c; }a[3005]; bool cmp(node x,node y) { return x.x<y.x; } ll dp[3005]; //dp[i]表示只考虑前i个教室 的最优解 ll dp2[3005]; //dp2[i]表示 第i个固定条件下，前i个的最优解 ll s[3005]; int main() { int n; while(~scanf("%d",&n)) { for(int i=1;i<=n;i++) { scanf("%lld %lld",&a[i].x,&a[i].c); } sort(a+1,a+n+1,cmp); memset(dp,inf,sizeof(dp)); memset(dp2,inf,sizeof(dp2)); memset(s,0,sizeof(s)); dp[0]=0; dp[1]=dp2[1]=a[1].c; int pp=a[1].x; for(int i=1;i<=n;i++) { a[i].x-=pp; s[i]=a[i].x+s[i-1]; } for(int i=1;i<=n;i++) { dp2[i]=dp[i-1]+a[i].c; //第i个固定条件下，前i个的最优解 for(int j=1;j<=i;j++) { dp[i]=min(dp[i],dp2[j]+s[i]-s[j]-(i-j)*a[j].x); } } printf("%lld\n",dp[n]); } return 0; }
Building Shops
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K
(Java/Others)
Total Submission(s): 2728 Accepted Submission(s):
936
The total cost consists of two parts. Building a candy shop at classroom i
Now Little Q wants to know how to build the candy shops with the minimal cost. Please write a program to help him.
In each test case, the first line contains an integer n(1≤n≤3000)
In the following n
There are no two classrooms having same coordinate.
