Building Shops

Building Shops

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

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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#include<iostream>#include<cstdio>#include<iomanip>#include<algorithm>#include<cstring>#include<string>using namespace std;#define ll long long int #define INF 0x3f3f3f3f#define maxsize 5050struct node {    ll x;    ll c;}a[maxsize];bool cmp(node a, node b){    return a.x < b.x;}ll dp[maxsize][maxsize];int n;int main(){    while (cin >> n)    {        for (int i = 1;i <= n;i++)        {            cin >> a[i].x >> a[i].c;        }        sort(a + 1, a + 1 + n, cmp);        for (int i=1;i <= n;i++)        {            dp[i][1] = dp[i][2] = INF;        }        dp[0][1] = 0;        dp[0][2] = 0;        for (int i = 1;i <= n;i++)        {            dp[i][1] = min(dp[i - 1][1], dp[i - 1][2]) + a[i].c;            ll sum = 0;                        for (int j = i - 1; j >= 1; j--)                            {                                sum = sum + (i - j)*(a[j + 1].x - a[j].x);                                dp[i][2] = min(dp[i][2], dp[j][1] + sum);                           }        }        cout << min(dp[n][1], dp[n][2]) << endl;    }    return 0;}