LOJ#2353 货币兑换

CDQ分治优化斜率优化DP。

有个结论就是每天买完卖完....知道这个之后考虑今天卖的是哪天买的就能写出n²DP了。

发现形式是f_i = max(a_ib_j + c_id_j)的形式。我们可以把c_i除出来，就是斜率优化了。

然后发现横坐标和斜率全部没有单调性，于是CDQ分治搞一搞。

#include <bits/stdc++.h>

const int N = 1000010;
const long double eps = 1e-12;

long double a[N], b[N], c[N], R[N], f[N], s, k[N], v[N], w[N];
int n, node[N], t[N], stk[N], top;

template <class T> inline void Max(T &a, const T &b) {
    a < b ? a = b : 0;
    return;
}

inline bool cmp_k(const int &a, const int &b) {
    return k[a] < k[b];
}

inline bool cmp_w(const int &a, const int &b) {
    return w[a] < w[b];
}

inline bool check(int a, int b, int c) {
    if((v[b] - v[a]) * (w[c] - w[b]) + eps >= (v[c] - v[b]) * (w[b] - w[a])) {
        return 1;
    }
    return 0;
}

void CDQ(int l, int r) {

    //printf("CDQ : l = %d r = %d 
", l, r);

    if(l == r) {
        if(r > 1) f[r] *= b[r];
        Max(f[r], f[r - 1]);
        v[r] = -f[r] / c[r];
        w[r] = -v[r] * R[r];
        //std::cout << "v[i] = " << v[r] << std::endl;
        //printf("f = %.3f 
", f[r]);
        return;
    }
    int mid = (l + r) >> 1;
    CDQ(l, mid);

    /// update
    memcpy(t + l, node + l, (r - l + 1) * sizeof(int));
    std::sort(t + l, t + mid + 1, cmp_w);
    std::sort(t + mid + 1, t + r + 1, cmp_k);
    // build convex
    stk[top = 1] = t[l];
    if(mid - l + 1 >= 2){
        stk[++top] = t[l + 1];
        for(int i = l + 2; i <= mid; i++) {
            while(top > 1 && check(stk[top - 1], stk[top], t[i])) {
                top--;
            }
            stk[++top] = t[i];
        }
    }
    // update f
    //printf("top = %d 
", top);
    int head = 1;
    for(int i = mid + 1; i <= r; i++) {
        int x = t[i];
        while(head < top && (v[stk[head]] - v[stk[head + 1]]) / (w[stk[head]] - w[stk[head + 1]]) + eps < k[x]) {
            head++;
        }
        int y = stk[head];
        //printf("Max %.3f  %.3f * %.3f 
", f[x], -v[y], (R[y] * a[x] + b[x]));
        //Max(f[x], -v[y] * (R[y] * a[x] + b[x]));
        Max(f[x], w[y] * k[x] - v[y]);
    }

    CDQ(mid + 1, r);
    return;
}

int main() {

    //freopen("in.in", "r", stdin);
    //freopen("my.out", "w", stdout);

    scanf("%d%Lf", &n, &s);
    for(int i = 1; i <= n; i++) {
        scanf("%Lf%Lf%Lf", &a[i], &b[i], &R[i]);
        c[i] = a[i] * R[i] + b[i];
        //std::cout << "c[i] = " << c[i] << std::endl;
        k[i] = a[i] / b[i];
        node[i] = i;
    }

    f[1] = s;

    CDQ(1, n);

    printf("%.3Lf
", f[n]);
    return 0;
}

eps很重要...