poj 3709 K-Anonymous Sequence 斜率優化dp 分組有大小限制

分組，並且每組的大小有限制

注意不一定要分滿組才最小，要取min

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;
const int MAXN=500010;
#define ll __int64

ll dp[MAXN];
ll a[MAXN];
int q[MAXN];//隊列
ll sum[MAXN];

int head,tail,n,m;

ll getDP(int i,int k)
{
    return dp[k]+sum[i]-sum[k]-a[k+1]*(i-k);
}

ll getUP(int k1,int k2) //yj-yk部分
{
    return (dp[k1]-sum[k1]+a[k1+1]*k1)-(dp[k2]-sum[k2]+a[k2+1]*k2);
}
ll getDOWN(int k1,int  k2)
{
    return a[k1+1]-a[k2+1];
}

int main()
{
	int T;
	scanf("%d",&T);
    while(T--)
    {
		scanf("%d%d",&n,&m);
		int i,j;
		sum[0]=0;
        for(i=1;i<=n;i++)
           scanf("%I64d",&a[i]);
		sort(a+1,a+n+1);
		for(i=1;i<=n;i++)
			sum[i]=sum[i-1]+a[i];
        head=tail=0;
		dp[0]=0;
        for(i=1;i<m*2&&i<=n;i++)
			dp[i]=getDP(i,0);
		q[tail++]=m;
        for(i=m*2;i<=n;i++)
        {
            //把斜率轉成相乘，注意順序，否則不等號方向會改變的
            while(head+1<tail &&  getUP(q[head+1],q[head])<=i*getDOWN(q[head+1],q[head]))
               head++;
            dp[i]=getDP(i,q[head]);

            while(head+1<tail && getUP(i-m+1,q[tail-1])*getDOWN(q[tail-1],q[tail-2])<=getUP(q[tail-1],q[tail-2])*getDOWN(i-m+1,q[tail-1]))
                    tail--;
            q[tail++]=i-m+1;
        }
        printf("%I64d\n",dp[n]);
    }
    return 0;
}