F - 三分
Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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.
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|<=10 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
4
0.6 5
3.9 10
5.1 7
8.4 10
Sample Output
Case #1: 832
题目大意：类似于回归曲线拟合，寻找 最优解。
思路分析：很明显从区间左->区间右，unhappiness是先减。。。。后增，目测单峰函数，用三分姿势
注意精度不要太高，弱卡1e-8，1950ms险过orz.
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int maxn=50000+100;
#define eps 1e-8
const int inf=0xfffffff;
const double pi=acos(-1.0);
struct nod
{
double x;
double w;
};
nod s[maxn];
int n;
double len(double a)
{
double sum=0;
for(int i=0;i<n;i++)
{
double d=fabs(s[i].x-a);
sum+=d*d*d*s[i].w;
}
return sum;
}
int kase=0;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
double l=inf,r=-inf;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&s[i].x,&s[i].w);
l=min(l,s[i].x);
r=max(r,s[i].x);
}
while(l+eps<=r)
{
double mid=(l+r)/2;
double mmid=(mid+r)/2;
double t1=len(mid),t2=len(mmid);
if(t1<=t2) r=mmid;
else l=mid;
}
printf("Case #%d: %.lf\n",++kase,len(l));
}
return 0;
}
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int maxn=50000+100;
#define eps 1e-8
const int inf=0xfffffff;
const double pi=acos(-1.0);
struct nod
{
double x;
double w;
};
nod s[maxn];
int n;
double len(double a)
{
double sum=0;
for(int i=0;i<n;i++)
{
double d=fabs(s[i].x-a);
sum+=d*d*d*s[i].w;
}
return sum;
}
int kase=0;
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
double l=inf,r=-inf;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&s[i].x,&s[i].w);
l=min(l,s[i].x);
r=max(r,s[i].x);
}
while(l+eps<=r)
{
double mid=(l+r)/2;
double mmid=(mid+r)/2;
double t1=len(mid),t2=len(mmid);
if(t1<=t2) r=mmid;
else l=mid;
}
printf("Case #%d: %.lf\n",++kase,len(l));
}
return 0;
}