POJ 3709 K-Anonymous Sequence

$dp$，斜率优化。

设$dp[i]$表示前$i$个位置调整成$K-Anonymous$的最小花费。

那么，$dp[i]=min(dp[j]+sum[i]-sum[j]-x[j+1]*(i-j))$。

直接算是$O(n^2)$，进行斜率优化即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-10;
void File()
{
freopen("D:\\in.txt","r",stdin);
freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
char c = getchar();
x = 0;
while(!isdigit(c)) c = getchar();
while(isdigit(c))
{
x = x * 10 + c - '0';
c = getchar();
}
}

int n,k,T;
long long x[500010],sum[500010],dp[500010];
int f1,f2,q[500010];

bool delete1(int a,int b,int c)
{
if(dp[b]-sum[b]+b*x[b+1]-c*x[b+1]<=
dp[a]-sum[a]+a*x[a+1]-c*x[a+1]
) return 1;
return 0;
}

bool delete2(int a,int b,int c)
{
if(
((dp[c]-sum[c]+c*x[c+1])-(dp[b]-sum[b]+b*x[b+1]))*(x[b+1]-x[a+1])<=
((dp[b]-sum[b]+b*x[b+1])-(dp[a]-sum[a]+a*x[a+1]))*(x[c+1]-x[b+1])
) return 1;
return 0;
}

int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
{
scanf("%lld",&x[i]);
sum[i]=sum[i-1]+x[i];
}

for(int i=k;i<2*k;i++) dp[i]=sum[i]-i*x[1];

f1=0; f2=1; q[0]=0; q[1]=k;

for(int i=2*k;i<=n;i++)
{
while(1)
{
if(f2-f1+1<2) break;
if(delete1(q[f1],q[f1+1],i)) f1++;
else break;
}

dp[i]=dp[q[f1]]+sum[i]-sum[q[f1]]-(i-q[f1])*x[q[f1]+1];

while(1)
{
if(f2-f1+1<2) break;
if(delete2(q[f2-1],q[f2],i-k+1)) f2--;
else break;
}

f2++;
q[f2]=i-k+1;
}

printf("%lld\n",dp
);
}
return 0;
}