HDU 6024 Building Shops（DP）

最新推荐文章于 2019-09-17 09:34:00 发布

转载最新推荐文章于 2019-09-17 09:34:00 发布 · 57 阅读

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原文链接：http://www.cnblogs.com/Archger/p/8451649.html

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本文介绍了一道经典的算法题目，即如何在一系列教室中选择若干位置建立糖果店以达到总成本最小化的目标。该问题涉及到动态规划算法，并提供了一个具体的实现方案。

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

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#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);

		}
		
}

转载于:https://www.cnblogs.com/Archger/p/8451649.html