HDU6024：Building Shops（简单DP）

Building Shops

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4446    Accepted Submission(s): 1551

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

 

 

题意：

在n个在一条线的教室里面开店，给出每个教室的位置和开店需要的钱，若该教室开店就消耗开店花的钱，不开店则消耗该教室到左边最近的开店的教室的距离的钱，不开店的教室左边一定有开店的教室，求最少花费多少钱。

题解：

由题意可得第一个教室一定会开店，由题目的两个状态类似背包，因此可以用DP来做，设DP[n][2]，DP[n][0]代表店铺n-1没有开店花费最少的钱，DP[0][1]代表店铺n-1开店花费最少的钱。

因此可以得一转移方程为dp[i][1] = classes[i].price + min(dp[i - 1][0], dp[i - 1][1]);。而关于dp[i][0]则可以枚举来得到最小值。

重点：m += (i - f) * (classes[f + 1].point - classes[f].point);

注意：该题数据应用long long！我因为这个卡了十分钟……

具体见如下代码：

#define _CRT_SECURE_NO_DepRECATE
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <iostream>
#include <cmath>
#include <iomanip>
#include <string>
#include <algorithm>
#include <bitset>
#include <cstdlib>
#include <cctype>
#include <iterator>
#include <vector>
#include <cstring>
#include <cassert>
#include <map>
#include <queue>
#include <set>
#include <stack>
#define ll long long
#define INF 0x3f3f3f3f
#define ld long double
const ld pi = acos(-1.0L), eps = 1e-8;
int qx[4] = { 0,0,1,-1 }, qy[4] = { 1,-1,0,0 }, qxx[2] = { 1,-1 }, qyy[2] = { 1,-1 };
using namespace std;
struct node
{
	ll point, price;
}classes[5000];
bool cmp(node x, node y)
{
	return x.point < y.point;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	ll dp[5000][2];
	while (cin >> n)
	{
		memset(dp, 0, sizeof(dp));
		for (int i = 0; i < n; i++)
		{
			cin >> classes[i].point >> classes[i].price;
		}
		sort(classes, classes + n, cmp);//题目没说按照顺序输入，所以需要先排序
		dp[0][1] = classes[0].price;
		dp[0][0] = INF;
		for (int i = 1; i < n; i++)
		{
			dp[i][1] = classes[i].price + min(dp[i - 1][0], dp[i - 1][1]);
			dp[i][0] = INF;
			ll m = 0;
			for (int f = i - 1; f >= 0; f--)//枚举出i店铺左边第一个店铺为哪个店铺的时候花费最少，从i-1开始枚举
			{
				m += (i - f) * (classes[f + 1].point - classes[f].point);//重点，可画图理解，m为每次往前推一点的时候增加的距离花费
				dp[i][0] = min(dp[i][0], dp[f][1] + m);
			}
		}
		cout << min(dp[n - 1][0], dp[n - 1][1]) << endl;
	}
	return 0;
}