[ZJOI2007]仓库建设

嘟嘟嘟


刚开始推了一个(O(n ^ 2))的dp方程，但是需要倒着来，然后斜率优化的时候出现了各种错误，最终还是放弃。
换一个正着来的吧。
令(dp[i])表示在(i)建仓库时的最小花费，则

[dp[i] = min _ {j = 0} ^ {i} {dp[j] + sum _ {k = j + 1} ^ {i} {p[k] * (x[i] - x[k])}} + c[i] ]

把里面的(sum)拆开，得到(x[i] * sum _ {k = j + 1} ^ {i} p[k] - sum _ {k = j + 1} ^ {i} d_k * p_k)，发现这个可以用两个前缀和优化到(O(1))。
于是

[dp[i] = min _ {j = 0} ^ {i} {dp[j] + x[i] * (sump[i] - sump[j]) - (sum[i] - sum[j])} + c[i] ]

然后就是正常的斜率优化了。
简单导一导，得到
(y = dp[j] + sum[j])
(k = x[i])
(x = sump[j])
(dp[i] = y - kx + d[i] * sump[i] - sum[i] + c[i])
发现(x, k)都是单调递增，很开心，维护下凸包即可。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e6 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, d[maxn], p[maxn], c[maxn];
ll sump[maxn], sum[maxn], dp[maxn];
int q[maxn], l = 0, r = 0;

#define k(i) d[i]
#define x(i) sump[i]
#define y(i) (dp[i] + sum[i])
db slope(int i, int j)
{
  return 1.0 * (y(i) - y(j)) / (x(i) - x(j));
}

int main()
{
  n = read();
  for(int i = 1; i <= n; ++i)
    {
      d[i] = read(), p[i] = read(), c[i] = read();
      sump[i] = sump[i - 1] + p[i];
      sum[i] = sum[i - 1] + (ll)d[i] * p[i];
    }
  for(int i = 1; i <= n; ++i)
    {
      while(l < r && slope(q[l], q[l + 1]) < k(i)) ++l;
      dp[i] = y(q[l]) - k(i) * x(q[l]) + d[i] * sump[i] - sum[i] + c[i];
      while(l < r && slope(q[r], q[r - 1]) > slope(q[r], i)) --r;
      q[++r] = i;
    }
  write(dp[n]), enter;
  return 0;
}