标签：names bit pen ++ dig print 状态转移方程 online www
[题目链接]
https://www.lydsy.com/JudgeOnline/problem.php?id=1911
[算法]
设前i个士兵"修正"后的最大战斗力为fi
令sumi表示x的前缀和
显然 ， 有状态转移方程 : fi = max{ fj + a * (sumi - sumj) ^ 2 + b * (sumi - sumj) + c }
对该式进行化简 ， 得 :
fi = max{ fj + asumi ^ 2 + asumj ^ 2 - 2asumisumj + bsumi - bsumj + c}
令Yj = fj + asumj ^ 2 , Xj = sumj
则 : fi = max{Yj - Xj(2sumi + b) + aumi ^ 2 + bsumi + c}
那么Yj = Xj + (2asumi + b) + fi - asumi ^ 2 - bsumi - c
显然我们要做的是最大化截距
2asumi + b单调递减 ， Xi单调递增 ， 维护一个上凸壳即可
时间复杂度 : O(N)
[代码]
#include<bits/stdc++.h> using namespace std; const int N = 1000010; typedef long long ll; typedef long double ld; typedef unsigned long long ull; int n , l , r; ll a , b , c; int q[N]; ll f[N] , sum[N] , X[N] , Y[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; } int main() { read(n); read(a); read(b); read(c); for (int i = 1; i <= n; i++) { int x; read(x); sum[i] = sum[i - 1] + x; X[i] = sum[i]; } f[q[l = r = 1] = 0] = 0; for (int i = 1; i <= n; i++) { while (l < r && Y[q[l + 1]] - Y[q[l]] >= (2 * a * sum[i] + b) * (X[q[l + 1]] - X[q[l]])) ++l; f[i] = Y[q[l]] - X[q[l]] * (2 * a * sum[i] + b) + a * sum[i] * sum[i] + b * sum[i] + c; Y[i] = f[i] + a * sum[i] * sum[i]; while (l < r && (Y[i] - Y[q[r]]) * (X[q[r]] - X[q[r - 1]]) >= (Y[q[r]] - Y[q[r - 1]]) * (X[i] - X[q[r]])) --r; q[++r] = i; } printf("%lld\n" , f[n]); return 0; }
标签：names bit pen ++ dig print 状态转移方程 online www
原文地址：https://www.cnblogs.com/evenbao/p/10354200.html
