HDU 5073 Galaxy（贪心）

题目链接：HDU 5073 Galaxy

题面：

Galaxy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2571    Accepted Submission(s): 642
Special Judge

Problem Description

Good news for us: to release the financial pressure, the government started selling galaxies and we can buy them from now on! The first one who bought a galaxy was Tianming Yun and he gave it to Xin Cheng as a present.


To be fashionable, DRD also bought himself a galaxy. He named it Rho Galaxy. There are n stars in Rho Galaxy, and they have the same weight, namely one unit weight, and a negligible volume. They initially lie in a line rotating around their center of mass.

Everything runs well except one thing. DRD thinks that the galaxy rotates too slow. As we know, to increase the angular speed with the same angular momentum, we have to decrease the moment of inertia.

The moment of inertia I of a set of n stars can be calculated with the formula


where w_i is the weight of star i, d_i is the distance form star i to the mass of center.

As DRD’s friend, ATM, who bought M78 Galaxy, wants to help him. ATM creates some black holes and white holes so that he can transport stars in a negligible time. After transportation, the n stars will also rotate around their new center of mass. Due to financial pressure, ATM can only transport at most k stars. Since volumes of the stars are negligible, two or more stars can be transported to the same position.

Now, you are supposed to calculate the minimum moment of inertia after transportation.

 

Input

The first line contains an integer T (T ≤ 10), denoting the number of the test cases.

For each test case, the first line contains two integers, n(1 ≤ n ≤ 50000) and k(0 ≤ k ≤ n), as mentioned above. The next line contains n integers representing the positions of the stars. The absolute values of positions will be no more than 50000.

 

Output

For each test case, output one real number in one line representing the minimum moment of inertia. Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.

 

Sample Input

23 2-1 0 14 2-2 -1 1 2

 

Sample Output

00.5

 

Source

2014 Asia AnShan Regional Contest

 

解题：

    重心的位置时所有位置的平均数，而能移动的点最后肯定都落在重心上。故只需找出哪些是保留的点，而保留的点应该是连续的，这样才能保证力矩之和最小。直接暴力是n^2，显然会超时。因为要求的是，∑从1到n(d^2)，而（di-dave）^2可以拆分成di^2-2*di*dave+dave^2可以先预处理di和di^2的累积和，最后算的时候直接取就行了。因为初值取太小了，导致wa多次。貌似是需要，排下序的。

总结：

1.不要认为题目给定的都是有序的。

2.求最值，初值应设为第一项或末项的值。

代码：

#include <iostream>#include <cmath>#include <algorithm>#include <cstdio>using namespace std;int main(){    int t,n,k;    long long int p[50000+10];    long long int d[50000+10];    long long int ds[50000+10];    double center;    double minn=0,temp;    cin>>t;    while(t--)    {        minn=9999999999999999;        cin>>n>>k;        for(int i=1;i<=n;i++)            cin>>p[i];        sort(p+1,p+n+1);        if(n==k||n-k==1)        {            printf("0\n");            continue;        }        d[0]=0;        ds[0]=0;        for(int i=1;i<=n;i++)        {            d[i]=d[i-1]+p[i];            ds[i]=ds[i-1]+p[i]*p[i];        }        for(int i=0;i<=k;i++)        {            center=(d[i+n-k]-d[i])*1.0/(n-k);            temp=(ds[i+n-k]-ds[i])-2*center*(d[i+n-k]-d[i])+(n-k)*center*center;            if(temp<minn)                minn=temp;        }        printf("%.9f\n",minn);    }    return 0;}