HDU 4355：Party All the Time（三分模板）

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6979 Accepted Submission(s): 2181

Problem Description

In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebrate the harvest every year. But there is one thing you may not know that they hate walking so much that they would prefer to stay at home if they need to walk a long way.According to our observation,a spirit weighing (W) will increase its unhappyness for (S^3*W) units if it walks a distance of S kilometers.
Now give you every spirit's weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.

Input

The first line of the input is the number (T(Tleq 20)), which is the number of cases followed. The first line of each case consists of one integer (N(1leq Nleq 50000)), indicating the number of spirits. Then comes (N) lines in the order that (x[i]leq x[i+1]) for all (i(1leq i<N)). The (i)-th line contains two real number : (X_i),(W_i), representing the location and the weight of the (i)-th spirit. ( (|x_i|leq10^6, 0<w_i<15) )

Output

For each test case, please output a line which is "Case #X: Y", X means the number of the test case and Y means the minimum sum of unhappyness which is rounded to the nearest integer.

Sample Input

1
4
0.6 5
3.9 10
5.1 7
8.4 10

Sample Output

Case #1: 832

题意

有(n)个精灵在一维坐标轴上，并且每个精灵都有一个权值(w)，每个精灵从一个点到达一个点要花费：(s^3*w)（(s)代表距离），问所有的精灵要聚在一起，最小花费是多少。

思路

三分模板题。

精灵聚集在一起需要的花费为：(sum left( left| x_{i}-x ight| ^{3} imes w_{i} ight))（(x)为聚集在一起的坐标，(x_i)为每个精灵的坐标）

对$ (x_{i}-x) ^{3}$求二阶导数，可以发现这是一个凸函数，可以用三分进行解决。

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
const double eps=1e-6;
using namespace std;
double x[maxn];
double w[maxn];
int n;
inline double times(double a)
{
    return a*a*a;
}
inline double mul(double place)
{
    double ans=0.0;
    for(register int i=0;i<n;i++)
        ans+=times(abs(place-x[i]))*w[i];
    return ans;
}
inline double sanfen(double l,double r)
{
    double mid,midr;
    while(abs(r-l)>eps)
    {
        mid=(l+r)/2;
        midr=(mid+r)/2;
        if(mul(mid)>mul(midr))
            l=mid;
        else
            r=midr;
    }
    return mul(l);
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("/home/wzy/in.txt", "r", stdin);
        freopen("/home/wzy/out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    int t;
    int _=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        double l,r;
        scanf("%lf%lf",&x[0],&w[0]);
        l=x[0];r=x[0];
        for(register int i=1;i<n;i++)
        {
            scanf("%lf%lf",&x[i],&w[i]);
            l=min(l,x[i]);
            r=max(l,x[i]);
        }
        double ans=sanfen(l,r);
        printf("Case #%d: %lld
",++_,(ll)floor(ans+0.5));
    }
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl;
    #endif
    return 0;
}