HDU6024 Building Shops 2017-05-07 18:33 30人阅读 评论(0) 收藏

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

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

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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 long
const 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;
}