bzoj1563: [NOI2009]诗人小G

f[i]表示前i句诗的最小不协调度，f[i]=min(0<=j<i){f[j]+|s[i]-s[j]+i-j-1-L|^p}

val(j,i)=|s[i]-s[j]+i-j-1-L|^p 满足四边形不等式

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long double LD;
int P;
LD quick_pow(LD A)
{
    LD ret=1;int p=P;
    while(p!=0)
    {
        if(p%2==1)ret*=A;
        A*=A;p/=2;
    }
    return ret;
}

int n,s[110000];LD L,f[110000];
LD myabs(LD x){return x<0?-x:x;}
LD val(int j,int i){return quick_pow(myabs(s[i]-s[j]+i-j-1-L));}
struct node
{
    int id,l,r;
    node(){}
    node(int ID,int L,int R){id=ID;l=L;r=R;}
}q[110000];int h,t;

char ss[3100000];
int main()
{
    freopen("poet.in","r",stdin);
    freopen("poet.out","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%Lf%d",&n,&L,&P);
        s[0]=0;
        for(int i=1;i<=n;i++)
            scanf("%s",ss+1), s[i]=s[i-1]+strlen(ss+1);
        
        h=1,t=0;q[++t]=node(0,0,n);
        for(int i=1;i<=n;i++)
        {
            if(q[h].r<i)h++;
            q[h].l=i;f[i]=f[q[h].id]+val(q[h].id,i);
            
            if(h>t||f[i]+val(i,n)<=f[q[t].id]+val(q[t].id,n))
            {
                while(h<=t&&f[i]+val(i,q[t].l)<=f[q[t].id]+val(q[t].id,q[t].l))t--;
                if(h>t)q[++t]=node(i,1,n);
                else
                {
                    int l=q[t].l,r=q[t].r,ans;
                    while(l<=r)
                    {
                        int mid=(l+r)/2;
                        if(f[i]+val(i,mid)>f[q[t].id]+val(q[t].id,mid))
                        {
                            ans=mid;
                            l=mid+1;
                        }
                        else r=mid-1;
                    }
                    q[t].r=ans;
                    q[++t]=node(i,ans+1,n);
                }
            }
        }
        if(f[n]>1e18)printf("Too hard to arrange
");
        else printf("%lld
",(long long)f[n]);
        printf("--------------------
");
    }
    return 0;
}