HDU 4355 - Party All the Time（三分）

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 S³*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(T<=20), which is the number of cases followed. The first line of each case consists of one integer N(1<=N<=50000), indicating the number of spirits. Then comes N lines in the order that x_[i]<=x_[i+1] for all i(1<=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|<=10⁶, 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

140.6 53.9 105.1 78.4 10

 

Sample Output

Case #1: 832

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题意：

求出直线上各点距离一个点 的 （S^3 + w）的总和最小。

思路：

三分。

题目给出的函数是凸函数，求最小值，使用三分。

之前不会，比赛跪惨T^T。

CODE：

#include <iostream>#include <cstdio>#include <cmath>using namespace std;typedef long long ll;const double esp = 1e-8;const int maxn = 50005;double d[maxn], w[maxn];int n;double cal(double x){    double dis, ans = 0.0;    for(int i = 0; i < n; ++i) {        dis = fabs(d[i] - x);        ans += (dis*dis*dis) * w[i];    }    return ans;}double solve(){    double l, r, mid, mmid;    l = -1000000.0;    r = 1000000.0;    while(r - l > esp) {        mid = (l + r) / 2.0;        mmid = (mid + r) / 2.0;        if(cal(mid) < cal(mmid)) r = mmid;        else l = mid;//        printf("%lf %lf\n", cal(mid), cal(mmid));//        mid = l + (r - l) / 3.0;//        mmid = r - (r - l) / 3.0;//        if(cal(mid) < cal(mmid)) r = mmid;//        else l = mid;    }    return cal(l);}int main(){//freopen("in", "r", stdin);    int T;    scanf("%d", &T);    int ca = 0;    while(T--) {    ca++;        scanf("%d", &n);        for(int i = 0; i < n; ++i) {            scanf("%lf %lf", &d[i], &w[i]);        }        double Ans = solve();        printf("Case #%d: %.0lf\n", ca, Ans);    }    return 0;}