洛谷P4027 [NOI2007]貨幣兌換(dp 斜率優化 cdq 二分)
阿新 • • 發佈:2019-01-01
題意
Sol
解題的關鍵是看到題目裡的提示。。。
設\(f[i]\)表示到第\(i\)天所持有軟妹幣的最大數量,顯然答案為\(max_{i = 1}^n f[i]\)
轉移為\(f_i = max(f_{i - 1}, A_i \frac{f_j R_j}{A_j R_j + B_j} + B_i \frac{f_j}{A_j R_j + B_j})\)
設\(y_i = \frac{f_j}{A_j R_j + B_j}, x_i = \frac{f_j R_j}{A_j R_j + B_j}\)
顯然可以斜率優化,也就是拿一條斜率為\(\frac{A_i}{B_i}\)
但是這裡的斜率和\(x\)都是不單調的。
按照老祖宗說的
\(x\)不單調cdq
斜率不單調二分凸包
然後xjb寫一寫就好了。我寫的cdq複雜度是\(O(nlog^2n)\)的,每次暴力建左側的凸包,然後在右邊二分,雖然很好寫,但是在BZOJ上成功T飛。。
看了下SovietPower大佬的部落格發現有nlogn的做法Orz,就是先按斜率排序,然後轉移的時候把下標分為\(<mid\)和\(>= mid\)的,直接用類似雙指標的東西掃就行了
#include<bits/stdc++.h> #define Pair pair<double, double> #define MP(x, y) make_pair(x, y) #define fi first #define se second //#define int long long #define LL long long #define db double #define Fin(x) {freopen(#x".in","r",stdin);} #define Fout(x) {freopen(#x".out","w",stdout);} using namespace std; const int MAXN = 1e6 + 10, mod = 1e9 + 7, INF = 1e9 + 10; const double eps = 1e-9; template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;} template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;} template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;} template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;} template <typename A> inline void debug(A a){cout << a << '\n';} template <typename A> inline LL sqr(A x){return 1ll * x * x;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, S; struct Sta { int id; db A, B, R, x, y, f; void Get() { x = f * R / (A * R + B); y = f / (A * R + B); } bool operator < (const Sta &rhs) const { return f < rhs.f; } }a[MAXN], st[MAXN]; vector<Pair> v; double GetK(Pair a, Pair b) { if((b.fi - a.fi) < eps) return INF; return (b.se - a.se) / (b.fi - a.fi); } void GetConvexHull(int l, int r) { v.clear(); for(int i = l; i <= r; i++) { double x = a[i].x, y = a[i].y; while(v.size() > 1 && ((GetK(v[v.size() - 1], MP(x, y)) > GetK(v[v.size() - 2], v[v.size() - 1])) )) v.pop_back(); v.push_back(MP(x, y)); } } int cnt = 0; db Find(int id, db k) { int l = 0, r = v.size() - 1, ans = 0; while(l <= r) { int mid = l + r >> 1; if((mid == 0) || (GetK(v[mid - 1], v[mid]) > k)) l = mid + 1, ans = mid; else r = mid - 1; } return a[id].A * v[ans].fi + a[id].B * v[ans].se; } db CDQ(int l, int r) { if(l == r) { int i = l; chmax(a[i].f, a[i - 1].f); chmax(a[i].f, a[i].f * ((a[i].A * a[i].R + a[i].B) / (a[i].A * a[i].R + a[i].B))); a[l].Get(); return a[i].f; } int mid = l + r >> 1; db lmx = CDQ(l, mid); GetConvexHull(l, mid); for(int i = mid + 1; i <= r; i++) chmax(a[i].f, max(lmx, Find(i, -a[i].A / a[i].B))); CDQ(mid + 1, r); int tl = l, tr = mid + 1, tot = tl - 1; while(tl <= mid || tr <= r) { if((tr > r) || (tl <= mid && a[tl].x < a[tr].x)) st[++tot] = a[tl++];//這裡要加上tl <= mid else st[++tot] = a[tr++]; } db rt = 0; for(int i = l; i <= r; i++) a[i] = st[i], chmax(rt, a[i].f); return rt; } signed main() { // freopen("a.in", "r", stdin); N = read(); S = read(); for(int i = 1; i <= N; i++) scanf("%lf %lf %lf", &a[i].A, &a[i].B, &a[i].R), a[i].id = i; a[0].f = S; CDQ(1, N); db ans = 0; for(int i = 1; i <= N; i++) chmax(ans, a[i].f); printf("%.3lf", ans); return 0; }