BZOJ #3156. 防御准备

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题意

1~n放城堡/木偶，在第i位放城堡的 \(\ cost_i\) 给出，放木偶的 \(\ cost_i = (j-i)\)(j为i右边第一个城堡)

问最小花费

题解

普通\(\theta{(n^2)}\)Dp会挂 \(\ n\leq 100000\)

需要斜率优化，为了方便，我们从左往右Dp

f[i] 为在这个点放城堡的最小花费(因为第n个必须放城堡)

$f_i=\min\limits_{j<i}{a_i+f_j+1+2+\dots+(i-j-1)} (j为i左边第一个城堡)$

$化简得 f_i=\min\limits_{j<i}{a_i+f_j+(i-j)\times(i-j-1)}$

$f_i=\min\limits_{j<i}{f_j+\frac{j(j-1)}{2}+ij+a_i+\frac{i(i-1)}{2}}$

$设x_j=j, y_j=f_j+\frac{j(j-1)}{2}$

然后斜率优化

调试记录

\(\ y_j\)找错

#include <cstdio>
#include <algorithm>
#include <cstring>
#define INF 0x3f3f3f3f
#define maxn 1000005
#define int long long

using namespace std;

int f[maxn], q[maxn], a[maxn];
int l, r, n, sum[maxn];

int Y(int id){ return f[id] + sum[id]; }
double slope(int id1, int id2){ return 1.0 * (Y(id2) - Y(id1)) / (id2 - id1); }

signed main(){
	scanf("%lld", &n); sum[0] = 0;
	for (int i = 1; i <= n; i++){
		scanf("%lld", &a[i]);
		sum[i] = sum[i - 1] + i;
	}
	l = 0, r = 0;
	
	for (int i = 1; i <= n; i++){
		while (l < r && slope(q[l], q[l + 1]) <= i) l++;
		int j = q[l];
		f[i] = Y(j) - i * j + a[i] + sum[i - 1];
		while (l < r && slope(q[r - 1], q[r]) >= slope(q[r], i)) r--;
		q[++r] = i;
	}
	printf("%lld\n", f[n]);
	return 0;
}