BZOJ1564 NOI2009二叉查找树（区间dp）

　　首先按数据值排序，那么连续一段区间的dfs序一定也是连续的。

　　将权值离散化，设f[i][j][k]为i到j区间内所有点的权值都>=k的最小代价，转移时枚举根考虑是否修改权值即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
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 75
int n,m,b[N],f[N][N][N],sum[N];
struct data
{
    int x,y,z;
    bool operator <(const data&a) const
    {
        return x<a.x;
    }
}a[N];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj1564.in","r",stdin);
    freopen("bzoj1564.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i].x=read();
    for (int i=;i<=n;i++) b[i]=a[i].y=read();
    for (int i=;i<=n;i++) a[i].z=read();
    sort(a+,a+n+);sort(b+,b+n+);
    for (int i=;i<=n;i++) a[i].y=lower_bound(b+,b+n+,a[i].y)-b;
    for (int i=;i<=n;i++) sum[i]=sum[i-]+a[i].z;
    memset(f,,sizeof(f));
    for (int i=;i<=n;i++)
    {
        for (int k=;k<=n;k++)
        f[i][i-][k]=;
        for (int k=;k<=a[i].y;k++)
        f[i][i][k]=a[i].z;
        for (int k=a[i].y+;k<=n;k++)
        f[i][i][k]=a[i].z+m;
    }
    for (int k=;k<=n;k++) f[n+][n][k]=;
    for (int k=;k<=n;k++)
        for (int i=;i<=n-k+;i++)
        {
            int j=i+k-;
            for (int root=i;root<=j;root++)
            {
                f[i][j][a[root].y]=min(f[i][j][a[root].y],f[i][root-][a[root].y]+f[root+][j][a[root].y]+sum[j]-sum[i-]);
                for (int d=;d<=n;d++)
                f[i][j][d]=min(f[i][j][d],f[i][root-][d]+f[root+][j][d]+sum[j]-sum[i-]+m);
            }
            for (int d=n;d>=;d--) f[i][j][d]=min(f[i][j][d],f[i][j][d+]);
        }
    cout<<f[][n][];
    return ;
}