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。

 1 #include <bits/stdc++.h>
 2 
 3 const int N = 1000010;
 4 const long double eps = 1e-12;
 5 
 6 long double a[N], b[N], c[N], R[N], f[N], s, k[N], v[N], w[N];
 7 int n, node[N], t[N], stk[N], top;
 8 
 9 template <class T> inline void Max(T &a, const T &b) {
10     a < b ? a = b : 0;
11     return;
12 }
13 
14 inline bool cmp_k(const int &a, const int &b) {
15     return k[a] < k[b];
16 }
17 
18 inline bool cmp_w(const int &a, const int &b) {
19     return w[a] < w[b];
20 }
21 
22 inline bool check(int a, int b, int c) {
23     if((v[b] - v[a]) * (w[c] - w[b]) + eps >= (v[c] - v[b]) * (w[b] - w[a])) {
24         return 1;
25     }
26     return 0;
27 }
28 
29 void CDQ(int l, int r) {
30 
31     //printf("CDQ : l = %d r = %d \n", l, r);
32 
33     if(l == r) {
34         if(r > 1) f[r] *= b[r];
35         Max(f[r], f[r - 1]);
36         v[r] = -f[r] / c[r];
37         w[r] = -v[r] * R[r];
38         //std::cout << "v[i] = " << v[r] << std::endl;
39         //printf("f = %.3f \n", f[r]);
40         return;
41     }
42     int mid = (l + r) >> 1;
43     CDQ(l, mid);
44 
45     /// update
46     memcpy(t + l, node + l, (r - l + 1) * sizeof(int));
47     std::sort(t + l, t + mid + 1, cmp_w);
48     std::sort(t + mid + 1, t + r + 1, cmp_k);
49     // build convex
50     stk[top = 1] = t[l];
51     if(mid - l + 1 >= 2){
52         stk[++top] = t[l + 1];
53         for(int i = l + 2; i <= mid; i++) {
54             while(top > 1 && check(stk[top - 1], stk[top], t[i])) {
55                 top--;
56             }
57             stk[++top] = t[i];
58         }
59     }
60     // update f
61     //printf("top = %d \n", top);
62     int head = 1;
63     for(int i = mid + 1; i <= r; i++) {
64         int x = t[i];
65         while(head < top && (v[stk[head]] - v[stk[head + 1]]) / (w[stk[head]] - w[stk[head + 1]]) + eps < k[x]) {
66             head++;
67         }
68         int y = stk[head];
69         //printf("Max %.3f  %.3f * %.3f \n", f[x], -v[y], (R[y] * a[x] + b[x]));
70         //Max(f[x], -v[y] * (R[y] * a[x] + b[x]));
71         Max(f[x], w[y] * k[x] - v[y]);
72     }
73 
74     CDQ(mid + 1, r);
75     return;
76 }
77 
78 int main() {
79 
80     //freopen("in.in", "r", stdin);
81     //freopen("my.out", "w", stdout);
82 
83     scanf("%d%Lf", &n, &s);
84     for(int i = 1; i <= n; i++) {
85         scanf("%Lf%Lf%Lf", &a[i], &b[i], &R[i]);
86         c[i] = a[i] * R[i] + b[i];
87         //std::cout << "c[i] = " << c[i] << std::endl;
88         k[i] = a[i] / b[i];
89         node[i] = i;
90     }
91 
92     f[1] = s;
93 
94     CDQ(1, n);
95 
96     printf("%.3Lf\n", f[n]);
97     return 0;
98 }