CodeForces 1083 E The Fair Nut and Rectangles 斜率优化DP

The Fair Nut and Rectangles

题意:有n个矩形,然后你可以选择k个矩形,选择一个矩形需要支付代价 ai, 问 总面积- 总支付代价 最大能是多少, 保证没有矩形套矩形。

题解:

sort 一下 只有  x 会递增  y 递减

然后 f[i] = f[j] + (x[i]-x[j])*y[i] - a[i]

f[j] = f[i] - x[i] * y[i] + x[j] * y[i] + a[i]

即 y = f[j], x = x[j], k = y[i],  b = f[i] - x[i] * y[i] + a[i] 

我们需要维护 f[i] 尽可能大, 所以我们维护一个上突壳就好了。

代码:

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e6 + 100;
struct Node{
    int x, y;
    LL a;
    bool operator < (const Node & z) const{
        return x < z.x;
    }
}A[N];
LL f[N];
int q[N];
int main(){
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i)
        scanf("%d%d%lld", &A[i].x, &A[i].y, &A[i].a);
    sort(A+1, A+1+n);
    int L = 1, R = 1;
    for(int i = 1; i <= n; ++i){
        while(L < R && f[q[L+1]]-f[q[L]]>= 1ll*A[i].y * ((A[q[L+1]].x - A[q[L]].x))) ++L;
        f[i] = f[q[L]] + (1ll*A[i].x-A[q[L]].x)*A[i].y - A[i].a;
        while(L < R && ((long double)f[q[R]]-f[q[R-1]]) * (A[i].x - A[q[R]].x) <= ((long double)f[i]-f[q[R]])  * ((A[q[R]].x - A[q[R-1]].x))) --R;
        q[++R] = i;
    }
    LL ans = 0;
    for(int i = 1; i <= n; ++i) ans = max(ans, f[i]);
    cout << ans << endl;
    return 0;
}
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原文地址:https://www.cnblogs.com/MingSD/p/10113420.html