【codeforces 429D】Tricky Function

【题目链接】:http://codeforces.com/problemset/problem/429/D

【题意】

给你n个数字;
让你求出一段区间[l,r]
使得
(r−l)2+(∑rl+1a[i])2最小

【题解】

求出前缀和数组sum[i];
可以发现,如果把数组的下标i作为第一维坐标(x),前缀和sum[i]作为第二维坐标(y);
所求的式子就是任意两点之间的距离平方;
问题转化成:已知平面上的n个点;
求最近的两个点之间的距离的平方;
这个可以用分治的方法搞出来;
(感觉就是个剪枝的暴力);
据说复杂度是N⋅log2N

【Number Of WA】

2

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0),cin.tie(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1e5+100;
const LL INF = 8e18 + 1;
//4e18
//8e18

struct abc{
    LL x,y;
};

LL a[N],sum[N];
abc b[N],c[N];
int n;

LL sqr(LL x){
    return x*x;
}

LL dis(abc a,abc b){
    LL temp = 0;
    temp += sqr(a.x-b.x);
    temp += sqr(a.y-b.y);
    return temp;
}

LL query(int l,int r){
    LL ret = INF;
    if (l>=r) return ret;
    if (l+1==r) return dis(b[l],b[r]);
    int m = (l+r)>>1,k = 0;
    LL t1 = query(l,m),t2 = query(m+1,r),temp;
    ret = min(t1,t2);
    rep2(i,m,l){
        temp = sqr(b[i].x-b[m].x);
        if (temp>ret) break;
        c[++k] = b[i];
    }
    rep1(i,m+1,r){
        temp = sqr(b[i].x-b[m].x);
        if (temp>ret) break;
        c[++k] = b[i];
    }
    sort(c+1,c+1+k,[&] (abc a,abc b) {return a.y<b.y;});
    rep1(i,1,k)
        rep1(j,i+1,k){
            temp = sqr(c[j].y-c[i].y);
            if (temp > ret) break;
            ret = min(ret,dis(c[i],c[j]));
        }
    return ret;
}

int main(){
    //Open();
    Close();
    cin >> n;
    rep1(i,1,n) cin >> a[i];
    rep1(i,1,n) sum[i] = sum[i-1] + a[i];
    rep1(i,1,n){
        b[i].x = i,b[i].y = sum[i];
    }
    sort(b+1,b+1+n,[&] (abc a,abc b) { return a.x < b.x;});
    cout << query(1,n) << endl;
    return 0;
}