BZOJ1492 [NOI2007]货币兑换Cash

CDQ分治的例题。。。

首先可以想到是动态规划，我们令f[i]表示第i天可以赚到的最大钱数，x[i]、y[i]表示f[i]取到最优的时候a、b券的最大值。

则f[i] = max(f[i - 1], a[i] * x[j] + b[i] * y[j]) (j  < i)

然后我们优化一下，对于

a * x[j] + b * y[j] < a * x[k] + b * y[k] 且 j < k  =>

a * (x[j] - x[k]) < b * (y[k] - y[j])  =>

a / b < - (y[j] - y[k]) / (x[j] - x[k])

于是我们发现可以斜率优化，维护上凸壳，结果发现x不单调，大家都很厉害会用平衡树维护，但是我就不会

然后就想到了CDQ分治什么的。。。

 

/**************************************************************
    Problem: 1492
    User: rausen
    Language: C++
    Result: Accepted
    Time:1420 ms
    Memory:12140 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <algorithm>
 
using namespace std;
typedef double lf;
const lf inf = 1e20;
const lf eps = 1e-9;
const int N = 1e5 + 5;
 
inline int sgn(lf x) {
    return fabs(x) < eps ? 0 : x < 0 ? -1 : 1; 
}
 
struct point {
    lf x, y, a, b, k, rate;
    int id;
     
    inline bool operator < (const point &p) const {
        return sgn(x - p.x) == 0 ? sgn(y - p.y) < 0 : sgn(x - p.x) < 0;
    }
} p[N], tmp[N];
 
inline bool cmp(point p, point q) {
    return p.k > q.k;
} 
 
lf f[N];
int n, top, s[N];
 
inline lf get(int a, int b) {
    if (!b) return -inf;
    return sgn(p[a].x - p[b].x) == 0 ? inf : (p[b].y - p[a].y) / (p[b].x - p[a].x);
}
 
void work(int l, int r) {
    if (l == r) {
        f[l] = max(f[l], f[l - 1]);
        p[l].y = f[l] / (p[l].a * p[l].rate + p[l].b);
        p[l].x = p[l].rate * p[l].y;
        return;
    }
    int i, j, l1, l2, mid = l + r >> 1;
    for (i = l, l1 = l, l2 = mid + 1; i <= r; ++i)
        if (p[i].id <= mid) tmp[l1++] = p[i];
        else tmp[l2++] = p[i];
    for (i = l; i <= r; ++i) p[i] = tmp[i];
     
    work(l, mid);
    for (top = 0, i = l; i <= mid; ++i) {
        while (top > 1 && sgn(get(s[top - 1], s[top]) - get(s[top], i)) < 0) --top;
        s[++top] = i;
    }
    s[++top] = 0;
    for (i = mid + 1, j = 1; i <= r; ++i) {
        while (j < top && sgn(p[i].k - get(s[j], s[j + 1])) < 0) ++j;
        f[p[i].id] = max(f[p[i].id], p[s[j]].x * p[i].a + p[s[j]].y * p[i].b);
    }
    work(mid + 1, r);
     
    for (i = l, l1 = l, l2 = mid + 1; i <= r; ++i)
        if ((p[l1] < p[l2] || l2 > r) && l1 <= mid) tmp[i] = p[l1++];
        else tmp[i] = p[l2++];
    for (i = l; i <= r; ++i) p[i] = tmp[i];
}
 
int main() {
    int i;
    scanf("%d%lf", &n, &f[0]);
    for (i = 1; i <= n; ++i) {
        scanf("%lf%lf%lf", &p[i].a, &p[i].b, &p[i].rate);
        p[i].k = -p[i].a / p[i].b, p[i].id = i;
    }
    sort(p + 1, p + n + 1, cmp);
    work(1, n);
    printf("%.3lf\n", f[n]);
    return 0;
}