codeforces 429D

题意：给定一个数组你个数的数组a，定义sum(i, j)表示sigma(a[i],...a[j]),以及另外一个函数f(i, j) = (i - j)^2 + sum(i+1, j)^2

        求最小的f(i, j)(i < j)

思路：变形一下f(i, j) = (i - j)^2 + (sum[j] - sum[i])^2

        那么把i看成x，sum[i]看成y，那就等价于求二维平面的最近点对吗。

        二维平面求最近点对有一个经典nlognlogn的分治算法。。

code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define x first
#define y second
const int Inf = 0x3fffffff;
pair<int, int> p[101010], tmp[101010];
int n;
ll dis(int x, int y, int x1, int y1){
     ll dx = x - x1, dy = y - y1;
     return dx * dx + dy * dy;
}

bool cmp(const pii& a, const pii& b){
     return a.y < b.y;
}

ll X, Y, k;
ll solve(const int l,const int r){
    if (l == r) return Inf;
    int m = (l + r) >> 1;
    ll d = solve(l, m), d1 = solve(m+1, r);
    d = min(d, d1);
    k = 0;
    X = p[m].x;
    for (int i = l; i <= r; ++i)
         if ((X-p[i].x) * (X-p[i].x) <= d) tmp[k++] = p[i];
    sort(tmp, tmp + k, cmp);
    for (int i = 0; i < k; ++i){
        Y = tmp[i].y;
        for (int j = i+1; j < k; ++j){
             if ((Y - tmp[j].y) * (Y - tmp[j].y) > d) break;
             d = min(d, dis(tmp[i].x, tmp[i].y, tmp[j].x, tmp[j].y));
        }
    } 
    return d;
}

void solve(){
    p[0] = make_pair(0, 0);
    int u;
    for (int i = 1; i <= n; ++i)
         scanf("%d", &u), p[i].x = i, p[i].y = p[i-1].y + u;
    ll ans = solve(1, n);
    cout << ans << endl;
}

int main(){
//    freopen("a.in", "r", stdin);
    while (scanf("%d", &n) != EOF){
         solve();
    } 
}