BZOJ1911 [Apio2010]特别行动队 【斜率优化】

1911: [Apio2010]特别行动队

Time Limit: 4 Sec  Memory Limit: 64 MB

Submit: 5005  Solved: 2455

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Description

Input

Output

Sample Input

4

-1 10 -20

2 2 3 4

Sample Output

9

HINT

一定要好好纪念一下QAQ，本蒟蒻第一次自己推出斜率优化dp

有点模糊惹。。将就一下【捂脸】

#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 int read(){
	int 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;
}
int n,a,b,c,q[maxn],head,tail;
LL sum[maxn],f[maxn];
inline double slope(int u,int v){
	double y1 = f[u] + a * sum[u] * sum[u] - b * sum[u],y2 = f[v] + a * sum[v] * sum[v] - b * sum[v];
	return (y1 - y2) / (sum[u] - sum[v]);
}
inline LL g(LL x){
	return a * x * x + b * x + c;
}
inline LL getf(int i,int j){
	return f[j] + g(sum[i] - sum[j]);
}
int main()
{
	n = read(); a = read(); b = read(); c = read();
	REP(i,n) sum[i] = sum[i - 1] + read();
	head = tail = 0;
	for (int i = 1; i <= n; i++){
		while (head < tail && slope(q[head],q[head + 1]) + eps > 2 * a * sum[i])
			head++;
		f[i] = getf(i,q[head]);
		while (head < tail && slope(q[tail],q[tail - 1]) < slope(i,q[tail]) + eps)
			tail--;
		q[++tail] = i;
	}
	cout << f[n] << endl;
	return 0;
}