[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1597

[算法]

首先將所有土地按長為第一關鍵字 ， 寬為第二關鍵字排序

顯然 ， 當i > j , 且yi >= yj時 ， 土地j沒有用 ， 不妨使用單調棧彈出所有沒有用的土地

用fi表示前i塊土地的最小經費

顯然 ， fi = min{ fj + aibj }

斜率優化即可

時間復雜度 : O(N)

[代碼]

#include<bits/stdc++.h>
using
 
 namespace std;
#define N 50010
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const ll inf = 1e18;

struct info
{
        ll x , y;
} a[N];

ll n , l , r , top;
ll f[N];
ll q[N] , X[N] , Y[N] , s[N];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template 
 
<typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1
 
) + c - ‘0‘;
    x *= f;
}
inline bool cmp(info a , info b)
{
        if (a.x != b.x) return a.x < b.x;
        else return a.y < b.y;
}

int main()
{
        
        read(n);
        for (int i = 1; i <= n; i++)
        {
                read(a[i].x);
                read(a[i].y);        
        }
        sort(a + 1 , a + n + 1 , cmp);
        for (int i = 1; i <= n; i++)
        {
                while (top > 0 && a[i].y >= a[s[top]].y) --top;
                s[++top] = i;        
        }
        for (int i = 0; i < top; i++)
                X[i] = -a[s[i + 1]].y;
        q[f[l = r = 1] = 0] = 0;
        for (int i = 1; i <= top; i++)
        {
                while (l < r && Y[q[l + 1]] - Y[q[l]] <= a[s[i]].x * (X[q[l + 1]] - X[q[l]])) ++l;
                f[i] = f[q[l]] - a[s[i]].x * X[q[l]];
                Y[i] = f[i];
                while (l < r && (Y[i] - Y[q[r]]) * (X[q[r]] - X[q[r - 1]]) <= (Y[q[r]] - Y[q[r - 1]]) * (X[i] - X[q[r]])) --r;
                q[++r] = i;
        }
        printf("%lld\n" , f[top]);
        
        return 0;
    
}

[USACO 2008 MAR] 土地購買