嘟嘟嘟

只要会决策单调性，这题就是练手的

首先按矩形长排序，这样只用考虑宽了。
然后很容易搞出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表达到\(O(1)\)。
打表发现决策单调。然后就是正常的优化了。
二分的时候需要注意当前队列非空。要不然会像我一样，不开氧气AC,开了RE2个点。

#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;
}