Codeforces Round #245 (Div. 1) 429D - Tricky Function 最近点对
D. Tricky Function
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
codeforces.com/problemset/problem/429/D
Description
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 Input
4
1 0 0 -1
Sample Output
1
HINT
题意
给你n个数,让你求最小的f(i,j)
f(i,j)=(j-i)^2+(sum[j]-sum[i])^2
其中sum表示前缀和
题解:
简单分析一下,俩平方,就是距离嘛
把所有点都变成(i,sum[i])然后就是找最近点对了,然后我们有几种做法:
1.科学的暴力加剪枝
2.最近点对问题
3.对每一个点进行二分 nlogn感觉非常科学
代码
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 2000001
#define mod 10007
#define eps 1e-5
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
ll x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
//************************************************************************************** struct Point
{
ll x;
ll y;
}point[maxn];
int n;
int tmpt[maxn]; bool cmpxy(const Point& a, const Point& b)
{
if (a.x != b.x)
return a.x < b.x;
return a.y < b.y;
} bool cmpy(const int& a, const int& b)
{
return point[a].y < point[b].y;
} ll dis2(int i, int j)
{
return (point[i].x - point[j].x) * (point[i].x - point[j].x)
+ (point[i].y - point[j].y) * (point[i].y - point[j].y);
} ll sqr(ll x)
{
return x * x;
} ll Closest_Pair(int left, int right)
{
ll d = infll;
if (left == right)
return d;
if (left + == right)
return dis2(left, right);
int mid = (left + right) >> ;
ll d1 = Closest_Pair(left, mid);
ll d2 = Closest_Pair(mid + , right);
d = min(d1, d2);
int i, j, k = ;
//分离出宽度为d的区间
for (i = left; i <= right; i++) {
if (sqr(point[mid].x - point[i].x) <= d)
tmpt[k++] = i;
}
sort(tmpt, tmpt + k, cmpy);
//线性扫描
for (i = ; i < k; i++) {
for (j = i + ; j < k && sqr(point[tmpt[j]].y - point[tmpt[i]].y) < d;
j++) {
ll d3 = dis2(tmpt[i], tmpt[j]);
if (d > d3)
d = d3;
}
}
return d;
} int main()
{
scanf("%d", &n);
ll sum = ;
for (int i = ; i < n; ++i)
{
int x;
scanf("%d", &x);
point[i].x = i;
sum += x;
point[i].y = sum;
}
cout << Closest_Pair(, n - ) << endl;
return ;
}
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