HDU 4355 Party All the Time（三分法搜索）

HDU 4355

思路：三分法求f(x)极值。

f(x)是指位置为x时的愤怒值之和，是一个三次函数，且存在极值点使f(x)最小。

code：

/*
* @author Novicer
* language : C++/C
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
#define INF 2147483647
#define cls(x) memset(x,0,sizeof(x))
#define rise(i,a,b) for(int i = a ; i <= b ; i++)
using namespace std;
const double eps(1e-8);
typedef long long lint;

const int maxn = 500000 + 5;
double x[maxn];
double w[maxn];
int n;

double f(double pos){
	double sum = 0.0;
	for(int i = 1 ; i <= n ; i++){
		sum += (double)pow(abs(pos - x[i]) , 3) * w[i];
	}
	return sum;
}

double ts(double L , double R){
	double ans = -1;
	for(int i = 1 ; i < 30 ; i++){
		double mid1 = (L+R)/2.0;
		double mid2 = (mid1 + R)/2.0;
		if(f(mid1) >= f(mid2))
			L = mid1;
		else
			R = mid2;
	}
	ans = L;
	return ans;
}
int main(){
//	freopen("input.txt","r",stdin);
	int t ; cin >> t; int kase = 1;
	while(t--){
		cin >> n;
		double left = 1e7 , right = -1e7;
		for(int i = 1 ; i <= n ; i++){
			scanf("%lf%lf",&x[i],&w[i]);
			left = min(left , x[i]);
			right = max(right , x[i]);
		}
		printf("Case #%d: %.0f\n",kase++ , f(ts(left,right)));

	}
	return 0;
}

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