BZOJ1563 NOI2009诗人小G（动态规划+决策单调性）

　　设f[i]为前i行的最小不协调度，转移枚举这一行从哪开始，显然有f[i]=min{f[j]+abs(s[i]-s[j]+i-j-1-m)^p}。大胆猜想有决策单调性就好了。证明看起来很麻烦，从略。注意需要全程long double。

#include<bits/stdc++.h>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 100010
#define inf 1000000000000000001ll
#define ll long long
#define ld long double
int T,n,m,p,a[N],from[N],stk[N],L[N],R[N],top;
ld f[N];
char s[N][];
void print(int n)
{
    if (n==) return;
    print(from[n]);
    for (int i=from[n]+;i<n;i++) printf("%s ",s[i]);
    puts(s[n]);
}
ld ksm(ld a,int k)
{
    ld s=;
    for (;k;k>>=,a*=a) if (k&) s*=a;
    return s;
}
ld calc(int i,int j){return f[j]+ksm(abs(a[i]-a[j]-m),p);}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("c.in","r",stdin);
    freopen("c.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        n=read(),m=read()+,p=read();
        for (int i=;i<=n;i++) scanf("%s",s[i]),a[i]=a[i-]+strlen(s[i]);
        for (int i=;i<=n;i++) f[i]=inf,a[i]+=i;
        stk[top=]=;L[]=,R[]=n;
        for (int i=;i<=n;i++)
        {
            int l=,r=top;
            while (l<=r)
            {
                int mid=l+r>>;
                if (R[mid]>=i) from[i]=stk[mid],r=mid-;
                else l=mid+;
            }
            f[i]=calc(i,from[i]);
            while (L[top]>i&&calc(L[top],i)<calc(L[top],stk[top])) top--;
            l=max(L[top],i+),r=R[top];int x=R[top]+;
            while (l<=r)
            {
                int mid=l+r>>;
                if (calc(mid,i)<calc(mid,stk[top])) x=mid,r=mid-;
                else l=mid+;
            }
            R[top]=x-;if (x<=n) stk[++top]=i,L[top]=x,R[top]=n;
        }
        if (f[n]<inf) printf(LL,(ll)f[n]),print(n);
        else puts("Too hard to arrange");
        for (int i=;i<=;i++) putchar('-');if (T) printf("\n");
    }
    return ;
}