hdoj 6024 Building Shops

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

 

参考了http://blog.csdn.net/STILLxjy/article/details/72514980

 

题意：题意有点坑，总让我觉得这是国人写的英文版，

x轴上有n个班级  主要问题就是建与不建，就是0和1的问题，即dp ，问总费用最少为多少

1：在第i个教室建超市，费用因为ci
2：没有建超市的教室的费5用为它和它左边最接近的超市的坐标之间的距离

 

我们设dp[i][0]表示在教室i不建超市时前i个教室的费用和的最小值；

dp[i][1]表示在教室i建超市时前i个教室的费用和的最小值

那么我们很快可以得出：dp[i][1]=min(dp[i-1][0],dp[i-1][1])+ci;

关于dp[i][0],我们可以想到枚举离教室i最近的超市j的位置，然后取所有情况的最小值就可以了。

关于dp[i][0],由于可以承受住时间复杂度为O(n^2)的算法，那么我们就可以想到枚举离教室i最近的超市j的位置，然后取所有情况的最小值就可以了。

假设左边最近超市为j，那么教室j+1~教室i都不能建超市，所以教室j+1~教室i的费用分别为他们的位置到教室j之间的距离了。当前dp[i][0] = dp[j][1] +( 教室j+1~教室i的费用)

如果我们暴力求解，那么时间复杂度会变成O(n^3)，会超时。但是我们会发现由于j是从大到小变化的，所以就可以用：t mp+= (i - j) * dp[j+1].x - dp[j].x);来记录教室j+1~教室i的费用和了。

关于 tmp += (i - j) * (dp[j+1].x - dp[j].x); 的解释：
比如我们要算x3 - x1 ， x2 - x1的sum,那么由于保证了x是升序排列的，所以sum = (x3 - x2) + 2 * (x2 - x1).

AC代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
typedef  long long ll;
const int maxn =3030;
ll dp[maxn][maxn]={0};
struct classroom{
    int x,c;
}a[maxn];
int cmp(classroom xx,classroom yy){
    return xx.x<yy.x; //升序排列
}
int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i=1;i<=n;i++)
            scanf("%d%d",&a[i].x,&a[i].c);
        sort(a+1,a+n+1,cmp);
        dp[1][0]=(ll)1<<60;
        dp[1][1]=a[1].c;
        for(int i=2;i<=n;i++){
            dp[i][1]=min(dp[i-1][0],dp[i-1][1])+a[i].c;
            ll tmp=0;
            dp[i][0]=(ll)1<<60;
            for(int j=i-1;j>0;j--){
                tmp=(ll)(a[j+1].x-a[j].x)*(i-j)+tmp;
                dp[i][0]=min(dp[i][0],dp[j][1]+tmp);
            }
        }
        printf("%lld
",min(dp[n][0],dp[n][1]));
    }
    return 0;
}
/*     <<
 1的2进制是
000…0001
(这里前面0的个数和int的位数有关，32位机器，gcc里有31个0)，左移2位之后变成:
000…0100,
也就是10进制的4，所以说左移1位相对于乘以2的n次方
 */