[USACO 2008 MAR] 土地购买

[题目链接]

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

[算法]

首先将所有土地按长为第一关键字 ， 宽为第二关键字排序

显然 ， 当i > j , 且yi >= yj时 ， 土地j没有用 ， 不妨使用单调栈弹出所有没有用的土地

用fi表示前i块土地的最小经费

显然 ， fi = min{ fj + aibj }

斜率优化即可

时间复杂度 : O(N)

[代码]

#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 = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    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 = ; i <= n; i++)
        {
                read(a[i].x);
                read(a[i].y);
        }
        sort(a +  , a + n +  , cmp);
        for (int i = ; i <= n; i++)
        {
                while (top >  && a[i].y >= a[s[top]].y) --top;
                s[++top] = i;
        }
        for (int i = ; i < top; i++)
                X[i] = -a[s[i + ]].y;
        q[f[l = r = ] = ] = ;
        for (int i = ; i <= top; i++)
        {
                while (l < r && Y[q[l + ]] - Y[q[l]] <= a[s[i]].x * (X[q[l + ]] - 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 - ]]) <= (Y[q[r]] - Y[q[r - ]]) * (X[i] - X[q[r]])) --r;
                q[++r] = i;
        }
        printf("%lld\n" , f[top]);
 
        return ;
 
}