HDU - 3507 Print Article【斜率优化】

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Time limit 3000 ms
Memory limit 65536 kB

Zero has an old printer that doesn’t work well sometimes. As it is antique, he still like to use it to print articles. But it is too old to work for a long time and it will certainly wear and tear, so Zero use a cost to evaluate this degree.
One day Zero want to print an article which has N words, and each word i has a cost Ci to be printed. Also, Zero know that print k words in one line will cost

M is a const number.
Now Zero want to know the minimum cost in order to arrange the article perfectly.

Input

There are many test cases. For each test case, There are two numbers N and M in the first line (0 ≤ n ≤ 500000, 0 ≤ M ≤ 1000). Then, there are N numbers in the next 2 to N + 1 lines. Input are terminated by EOF.

Output

A single number, meaning the mininum cost to print the article.

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题目分析

好久没写斜率优化了，找了道水题练练手

$dp[i]$表示打印前 $i$个单词的最小花费
$dp[i]=dp[j]+(sum[i]-sum[j])^2+M$其中 $sum$为 $c_i$的前缀和

移向变成 $dp[j]+sum[j]^2=2*sum[i]*sum[j]-sum[i]^2-M+dp[i]$
按照斜率优化的套路维护下凸壳即可

这题好像很卡精度啊，不把斜率计算变成乘法一直WA =_=

---

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
typedef long long lt;
typedef double dd;
#define sqr(x) ((x)*(x))

int read()
{
    int f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();}
    while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();}
    return f*x;
}

const int maxn=500010;
int n,M;
lt sum[maxn],dp[maxn];
int q[maxn],ll,rr;

lt qX(int j){ return sum[j];}
lt qY(int j){ return dp[j]+sqr(sum[j]);}

lt query(int i,int j){ 
	return dp[j]+sqr(sum[i]-sum[j])+M;
}

int main()
{
    while(scanf("%d%d",&n,&M)!=EOF)
	{
		for(int i=1;i<=n;++i) 
		sum[i]=read()+sum[i-1];
		
		ll=rr=1;
		for(int i=1;i<=n;++i)
		{
			while( ll<rr&& (qY(q[ll+1])-qY(q[ll]))<=2*sum[i]*(qX(q[ll+1])-qX(q[ll])) ) ++ll;
			dp[i]=query(i,q[ll]);
			while(ll<rr&& (qY(q[rr])-qY(q[rr-1]))*(qX(i)-qX(q[rr]))>=(qY(i)-qY(q[rr]))*(qX(q[rr])-qX(q[rr-1])) ) --rr;
			q[++rr]=i;
		}
	
		printf("%d\n",dp[n]);
	}
	return 0;
}