HDU6024：Building Shops（簡單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中國大學生程序設計競賽 - 女生專場

 

 

題意：

在n個在一條線的教室裏面開店，給出每一個教室的位置和開店須要的錢，若該教室開店就消耗開店花的錢，不開店則消耗該教室到左邊最近的開店的教室的距離的錢，不開店的教室左邊必定有開店的教室，求最少花費多少錢。

題解：

由題意可得第一個教室必定會開店，由題目的兩個狀態相似揹包，所以能夠用DP來作，設DP[n][2]，DP[n][0]表明店鋪n-1沒有開店花費最少的錢，DP[0][1]表明店鋪n-1開店花費最少的錢。

所以能夠得一轉移方程爲dp[i][1] = classes[i].price + min(dp[i - 1][0], dp[i - 1][1]);。而關於dp[i][0]則能夠枚舉來獲得最小值。

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

注意：該題數據應用long long！我由於這個卡了十分鐘……

具體見以下代碼：

#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);//題目沒說按照順序輸入，因此須要先排序
		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--)//枚舉出i店鋪左邊第一個店鋪爲哪一個店鋪的時候花費最少，從i-1開始枚舉
			{
				m += (i - f) * (classes[f + 1].point - classes[f].point);//重點，可畫圖理解，m爲每次往前推一點的時候增長的距離花費
				dp[i][0] = min(dp[i][0], dp[f][1] + m);
			}
		}
		cout << min(dp[n - 1][0], dp[n - 1][1]) << endl;
	}
	return 0;
}