BZOJ 1492 [NOI 2007] 货币兑换Cash （dp + 分治）

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题目链接：BZOJ 1492

其实这种用单调队列来更新答案的dp可以用平衡树这种鬼畜做法来维护，做到时间复杂度为NlogN，前不久我还写了一道用splay维护的dp题。这道题就是学习一下cdq分治，代码确实比用splay写的要短。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;const int maxn = 100000 + 10;int N, S, top = 0, Left = 0, Right = 0;double eps = 1e-9;int q[maxn];double f[maxn];struct node{int id;double a, b, rate, k, x, y;}D[maxn], t[maxn];inline int read(){int x = 0, f = 1; char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1; ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0'; ch = getchar();}return x * f;}void input(){N = read(); f[0] = read();for(int i = 1; i <= N; ++i){scanf("%lf%lf%lf", &D[i].a, &D[i].b, &D[i].rate);D[i].k = - D[i].a / D[i].b; D[i].id = i;}}bool cmp(node A, node B){return A.k > B.k;}double get(int A, int B){if(!A)return -1e20;if(abs(D[A].x - D[B].x) < eps)return 1e20;return (D[A].y - D[B].y) / (D[A].x - D[B].x);}void cdq(int l, int r){if(l == r){f[l] = max(f[l], f[l - 1]);D[l].y = f[l] / (D[l].a * D[l].rate + D[l].b);D[l].x = D[l].y * D[l].rate;return ;}int mid = (l + r) >> 1;int L = l, R = mid + 1;for(int i = l; i <= r; ++i){if(D[i].id <= mid)t[L ++] = D[i];else t[R ++] = D[i];}for(int i = l; i <= r; ++i)D[i] = t[i];cdq(l, mid);Left = 1; Right = 0;for(int i = l; i <= mid; ++i){while(Right > 1 && get(q[Right], q[Right - 1]) < get(i, q[Right - 1]) + eps)Right --;q[++Right] = i;}q[++Right] = 0;for(int i = mid + 1; i <= r; ++i){while(Left < Right && get(q[Left + 1], q[Left]) + eps > D[i].k)Left ++;f[D[i].id] = max(f[D[i].id], D[q[Left]].x * D[i].a + D[q[Left]].y * D[i].b);}cdq(mid + 1, r);L = l; R = mid + 1;for(int i = l; i <= r; ++i){if((D[L].x < D[R].x || (abs(D[L].x - D[R].x) < eps && D[L].y < D[R].y) || (R > r)) && L <= mid)t[i] = D[L ++];else t[i] = D[R ++];}for(int i = l; i <= r; ++i)D[i] = t[i];}void solve(){sort(D + 1, D + N + 1, cmp);cdq(1, N);printf("%.3lf\n", f[N]);}int main(){input();solve();return 0;}

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