HDU 6024 Building Shops（DP）

Building Shops

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 889    Accepted Submission(s): 343

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

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<ctime>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
#define pi 3.14159265358979
#define MOD 1000000007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 3010;

struct Node {
	ll x, c;
}a[maxn];

ll dp[maxn][maxn];

bool cmp(Node a, Node b)
{
	return a.x < b.x;
}

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 + 1 + n, cmp);
			ms(dp, INF);
			dp[1][1] = a[1].c;
			for (int i = 2; i <= n; i++)
			{
				for (int j = 1; j < i; j++)
				{
					dp[i][i] = min(dp[i][i], dp[i - 1][j] + a[i].c);
					dp[i][j] = min(dp[i][j], dp[i - 1][j] + (a[i].x - a[j].x));
				}
			}
			ll ans = INF;
			for (int i = 1; i <= n; i++)
			{
				ans = min(ans, dp[n][i]);
			}
			printf("%lld\n", ans);

		}
		
}