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：
读了好久才明白
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：没有建糖果屋的教室的费用为它与左边离它最近的糖果屋的距离
解题思路：
从左到右，只需要O(n*n)，遍历一遍，每次只考虑当前教室为最后一个教室然后考虑局部状态，每一个教室只有两种状态，要么建成糖果屋，要么不建。跟背包思路相似，要么放进，要么不放。一个教室 i 建成糖果屋时费用为min(dp[i-1][1],dp[i-1][0])+建成它的费用。教室不建糖果屋费用为它到最近的糖果屋的距离，此时需要枚举在教室 i 前哪个教室建成糖果屋能够使得 i 不建糖果屋时花费最小（因为它前面的第 j 个教室虽然两者距离远，但是建成 j 糖果屋的花费小，所以要枚举）。但每次枚举都要考虑 i到j+1的教室到 j 的距离，复杂度又变成了O(n*n*n)，此时只需要 sum+=(i-j)*(a[j+1].first-a[j].first)此公式即可;
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define Max 3010
typedef pair<ll ,ll > p;
p a[Max];
ll dp[Max][2];
bool cmp(p x,p y)
{
return x.first<y.first;
}
int main(){
int n;
while(~scanf("%d",&n))
{
for(int i=0;i<n;i++)
scanf("%lld%lld",&a[i].first,&a[i].second);
sort(a,a+n,cmp);
dp[0][0]=INF;
dp[0][1]=a[0].second;
for(int i=1;i<n;i++)
{
ll sum=0;
dp[i][1]=min(dp[i-1][0],dp[i-1][1])+a[i].second;
dp[i][0]=INF;
for(int j=i-1;j>=0;j--)
{
sum+=(i-j)*(a[j+1].first-a[j].first);
dp[i][0]=min(dp[i][0],dp[j][1]+sum);
}
}
printf("%lld\n",min(dp[n-1][0],dp[n-1][1]));
}
return 0;
}