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;

}

巴特西

HDU6024：Building Shops（简单DP）

Building Shops

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