嘟嘟嘟


只要會決策單調性，這題就是練手的


首先按矩形長排序，這樣只用考慮寬了。
然後很容易搞出dp方程
\[dp[i] = min _ {j = 0} ^ {i - 1} (dp[j] + x[i] * max_{k = j + 1} ^ {i} y[k])\]
找max可以用st表達到\(O(1)\)。
打表發現決策單調。然後就是正常的優化了。
二分的時候需要註意當前隊列非空。要不然會像我一樣，不開氧氣AC,開了RE2個點。

#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 = 5e4 + 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;
ll dp[maxn];
struct Node
{
  ll x, y;
  bool operator < (const Node& oth)const
  {
    return x < oth.x || (x == oth.x && y < oth.y);
  }
}t[maxn];
struct Node2
{
  int pos, L, R;
}q[maxn];
int l = 1, r = 0;

ll Max[20][maxn], b[maxn];
void init()
{
  for(int i = 1; i <= n; ++i) Max[0][i] = t[i].y;
  for(int j = 1; (1 << j) <= n; ++j)
    for(int i = 1; i + (1 << j) - 1 <= n; ++i)
      Max[j][i] = max(Max[j - 1][i], Max[j - 1][i + (1 << (j - 1))]);
  int x = 0;
  for(int i = 1; i <= n; ++i)
    {
      if((1 << (x + 1)) <= i) x++;
      b[i] = x;
    }
}
ll query(int L, int R)
{
  int k = b[R - L + 1];
  return max(Max[k][L], Max[k][R - (1 << k) + 1]);
}

ll w(int L, int R)
{
  return query(L, R) * t[R].x;
}

int solve(int x, Node2 a)
{
  int L = a.L, R = a.R;
  while(L <= R)
    {
      int mid = (L + R) >> 1;
      if(dp[x] + w(x + 1, mid) <= dp[a.pos] + w(a.pos + 1, mid))
    {
      if(R != mid) R = mid;
      else {L = mid; break;}
    }
      else
    {
      if(L != mid + 1) L = mid + 1;
      else break;
    }
    }
  return L;
}

int main()
{
  n = read();
  for(int i = 1; i <= n; ++i) t[i].x = read(), t[i].y = read();
  sort(t + 1, t + n + 1);
  init();
  q[++r] = (Node2){0, 1, n};
  for(int i = 1; i <= n; ++i)
    {
      while(q[l].R < i) l++;
      dp[i] = dp[q[l].pos] + w(q[l].pos + 1, i);
      q[l].L = i + 1;
      while(l <= r && dp[i] + w(i + 1, q[r].L) <= dp[q[r].pos] + w(q[r].pos + 1, q[r].L)) r--;
      int pos = i;
      if(l <= r) pos = solve(i, q[r]), q[r].R = pos - 1;
      if(pos <= n) q[++r] = (Node2){i, pos, n};
    }
  write(dp[n]), enter;
  return 0;
}
 

[USACO08MAR]Land Acquisition