标签：复杂度 binary while char def problem target search 算法
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=2726
[算法]
此题与POJ1180非常相似
但是 ， 此题中的t值可能为负 ， 这意味着不能每次都将斜率 <= k的点弹出 ， 而需要在凸壳中进行二分查找
时间复杂度 : O(NlogN)
[代码]
#include<bits/stdc++.h> using namespace std; const int N = 1e6 + 10; typedef long long ll; typedef long double ld; typedef unsigned long long ull; int n , S , l , r; ll sumc[N] , sumt[N] , f[N]; int q[N] , c[N] , t[N]; template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); } template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); } template <typename T> inline void read(T &x) { T f = 1; x = 0; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f; for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘; x *= f; } inline int _binary_search(int L , int R , int val) { int l = L , r = R , ret = L; while (l <= r) { int mid = (l + r) >> 1; if (f[q[mid + 1]] - f[q[mid]] <= val * (sumc[q[mid + 1]] - sumc[q[mid]])) { ret = mid + 1; l = mid + 1; } else r = mid - 1; } return ret; } int main() { read(n); read(S); for (int i = 1; i <= n; i++) { read(t[i]); read(c[i]); sumt[i] = sumt[i - 1] + t[i]; sumc[i] = sumc[i - 1] + c[i]; } f[q[l = r = 1] = 0] = 0; for (int i = 1; i <= n; i++) { int pos = _binary_search(l , r - 1 , S + sumt[i]); f[i] = f[q[pos]] - sumc[q[pos]] * (S + sumt[i]) + sumt[i] * sumc[i] + S * sumc[n]; while (l < r && (f[i] - f[q[r]]) * (sumc[q[r]] - sumc[q[r - 1]]) <= (f[q[r]] - f[q[r - 1]]) * (sumc[i] - sumc[q[r]])) --r; q[++r] = i; } printf("%lld\n" , f[n]); return 0; }
标签：复杂度 binary while char def problem target search 算法
原文地址：https://www.cnblogs.com/evenbao/p/10354184.html
