HDU6024 Building Shops

Building Shops

                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                  Total Submission(s): 13    Accepted Submission(s): 7

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

31 22 33 441 73 15 106 1

 

Sample Output

511

 

Source

2017中国大学生程序设计竞赛 - 女生专场

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题意：给出n个地点的位置坐标xi和花费ci，对于每一个位置可以选择在花费ci建店或者花费左边离他最近的已建的位置的距离差，求最小花费

解题思路：dp，第一个点必建， dp[i][0]表示在该点不建，第一个点到第i个点的最小花费和，dp[i][1]表示该点建，第一个点到第i个点的最小花费和

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;int n;struct node{    LL x,c;} p[10004];LL dp[10004][2];bool cmp(node a,node b){    return a.x<b.x;}int main(){    while(~scanf("%d",&n))    {        memset(dp,INF,sizeof dp);        memset(p,0,sizeof p);        for(int i=0; i<n; i++)            scanf("%lld%lld",&p[i].x,&p[i].c);        sort(p,p+n,cmp);        dp[0][1]=p[0].c;        for(int i=1; i<n; i++)        {            dp[i][1]=min(dp[i-1][0],dp[i-1][1])+p[i].c;            LL dis=0;            for(int j=i-1; j>=0; j--)            {                dis+=(p[j+1].x-p[j].x)*(i-j);                dp[i][0]=min(dp[i][0],dp[j][1]+dis);            }        }        printf("%lld\n",min(dp[n-1][0],dp[n-1][1]));    }    return 0;}