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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[n]);    }    return 0;}

 

POJ 3709 K-Anonymous Sequence