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[], tmp[];
 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) >> ;
     ll d = solve(l, m), d1 = solve(m+, r);
     d = min(d, d1);
     k = ;
     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 = ; i < k; ++i){
         Y = tmp[i].y;
         for (int j = i+; 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[] = make_pair(, );
     int u;
     for (int i = ; i <= n; ++i)
          scanf("%d", &u), p[i].x = i, p[i].y = p[i-].y + u;
     ll ans = solve(, n);
     cout << ans << endl;
 }

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