BZOJ3156: 防御准备 【斜率优化dp】

3156: 防御准备

Time Limit: 10 Sec  Memory Limit: 512 MB
Submit: 2207  Solved: 933
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Description

Input

第一行为一个整数N表示战线的总长度。

第二行N个整数，第i个整数表示在位置i放置守卫塔的花费Ai。

Output

共一个整数，表示最小的战线花费值。

Sample Input

10
2 3 1 5 4 5 6 3 1 2

Sample Output

18

HINT

1<=N<=10^6,1<=Ai<=10^9

练了几题，这类题目的模式还是很固定好写的

就不推了

设f[i]表示i位置放防御塔的最小代价

显然有f[i] = min{f[j] + (i - j) * (i - j - 1) / 2} + A[i]    【中间那一段就是中间木偶的代价】

去掉常数化简有2 * i * j + 2 * f[i] = (2 * f[j] + j^2 + j)

令y = 2 * f[j] + j^2 + j，x = j

就是y = 2i * x + 2 * f[i]

化为求截距最小，由于所有值都是单调递增，维护下凸包

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define eps 1e-9
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define fo(i,x,y) for (int i = (x); i <= (y); i++)
#define Redge(u) for (int k = head[u]; k != -1; k = edge[k].next)
using namespace std;
const int maxn = 1000005,maxm = 100005,INF = 1000000000;
inline LL read(){
	LL out = 0,flag = 1;char c = getchar();
	while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57) {out = out * 10 + c - 48; c = getchar();}
	return out * flag;
}
LL n,q[maxn],head,tail;
LL A[maxn],f[maxn];
inline double slope(LL u,LL v){
	double y1 = 2 * f[u] + u * u + u,y2 = 2 * f[v] + v * v + v;
	return (y1 - y2) / (u - v);
}
inline LL getf(LL i,LL j){
	return f[j] + (i - j) * (i - j - 1) / 2 + A[i];
}
int main()
{
	n = read();
	REP(i,n) A[i] = read();
	head = tail = 0;
	for (LL i = 1; i <= n; i++){
		while (head < tail && slope(q[head],q[head + 1]) + eps < 2 * i) head++;
		f[i] = getf(i,q[head]);
		while (head < tail && slope(q[tail],q[tail - 1]) + eps > slope(i,q[tail])) tail--;
		q[++tail] = i;
	}
	cout<<f[n]<<endl;
	return 0;
}