Codeforces(429D - Tricky Function)最近点对问题

D. Tricky Function

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Iahub and Sorin are the best competitive programmers in their town. However, they can't both qualify to an important contest. The selection will be made with the help of a single problem. Blatnatalag, a friend of Iahub, managed to get hold of the problem before
the contest. Because he wants to make sure Iahub will be the one qualified, he tells Iahub the following task.

You're given an (1-based) array a with n elements.
Let's define function f(i, j) (1 ≤ i, j ≤ n) as (i - j)2 + g(i, j)2.
Function g is calculated by the following pseudo-code:

int g(int i, int j) {
int sum = 0;
for (int k = min(i, j) + 1; k <= max(i, j); k = k + 1)
sum = sum + a[k];
return sum;
}

Find a value mini ≠ j f(i, j).

Probably by now Iahub already figured out the solution to this problem. Can you?

Input

The first line of input contains a single integer n (2 ≤ n ≤ 100000).
Next line contains n integers a[1], a[2],
..., a[n] ( - 104 ≤ a[i] ≤ 104).

Output

Output a single integer — the value of mini ≠ j f(i, j).

Sample test(s)

input

4
1 0 0 -1

output

1

input

2
1 -1

output

2

解法：可以将结果转化为求(i，sum(i))最近点对问题。sum(i)为前缀1-i之和；

代码：

/******************************************************
* author:xiefubao
*******************************************************/
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <vector>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <string.h>
//freopen ("in.txt" , "r" , stdin);
using namespace std;

#define eps 1e-8
const double pi=acos(-1.0);
typedef long long LL;
const int Max=10100;
const int INF=1000000007;

struct point
{
double x,y;
int lable;
} ;
point points[1001000];

bool operator<(const point& a,const point& b)
{
if(a.x!=b.x)
return a.x<b.x;
else
return a.y<b.y;
}
bool compareY(const point& a,const point& b)
{
return a.y<b.y;
}
double getDistance(const point& a,const point& b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double getMiniDistance(int left,int right)
{
if(left==right)
return 1000000000000;
if(right-left==1)
{
if(points[left].lable^points[right].lable)
return getDistance(points[left],points[right]);
else
return 1000000000000;
}
int mid=(left+right)/2;
double num=min(getMiniDistance(left,mid),getMiniDistance(mid+1,right));
double mLine=points[mid].x;
int L=mid;
while(L>left&&mLine-points[L].x<=num)
L--;
int R=mid+1;
while(R<right&&points[R].x-mLine<=num)
R++;
sort(points+L,points+R+1,compareY);
for(int i=L; i<=R; i++)
{
for(int j=i+1; j<=min(R,i+5); j++)
{
if(points[j].y-points[i].y>=num)
break;
if(points[j].lable^points[i].lable)
{
num=min(num,getDistance(points[i],points[j]));
}
}
}
return num;
}

int main()
{
int T;
scanf("%d",&T);
for(int i=0; i<T; i++)
{
int N;
scanf("%d",&N);
for(int i=0; i<N; i++)
{
scanf("%lf%lf",&points[i].x,&points[i].y);
points[i].lable=0;
}
for(int i=N; i<N*2; i++)
{
scanf("%lf%lf",&points[i].x,&points[i].y);
points[i].lable=1;
}
sort(points,points+2*N);
printf("%.3f\n",getMiniDistance(0,2*N-1));
}
return 0;
}