C3-Zexal的OBST

题目描述

假设给定一个n个不同关键字的严格升序序列K=<k[1], k[2], …, k[n]>，用这些关键字构造二叉搜索树。对关键字k[i]，有p[i]次被检索到。有些搜索的值可能不在K中，假设n+1个伪关键字D=<d[0], d[1], …, d[n]>，对i=1, 2, ..., n-1，d[i]表示在k[i]和k[i+1]之间的值，d[0]表示小于k[1]的值，d[n]表示大于k[n]的值。对每个伪关键字d[i]，有q[i]次被检索到。请注意这里规定了每个关键字和伪关键字的检索次数。

用上述D序列作叶节点，K序列做内部节点，（可以参考算法导论第三版中文版226-230页，但注意题目定义的不同之处）构造一棵最优二叉搜索树。假设根节点深度为1，给定p, q，求出这二叉搜索棵树的最小总代价。

总代价定义为下面两式之和：

输入

第一行两个整数n。1≤n≤5001≤n≤500

第二行n个整数 p[i]p[i]，表示关键字的出现次数。

第三行n+1个整数q[i]q[i]，表示i-1关键字与i关键字之间的出现次数。0≤p[i],q[i]≤10000≤p[i],q[i]≤1000

输出

一个整数，表示最小总代价。

输入样例

5
15 10 5 10 20
5 10 5 5 5 10

输出样例

275

代码

#include <iostream>
#include <iomanip>
#include <cstring>
#include<cstdio>
#define N 505
using namespace std;
typedef int T;
int a[N],b[N];
T w[N][N],q[N],p[N],e[N][N];
int r[N][N];
int n;
T dep;
void PRINT_OPTIMAL_BST(int i,int j,T depth)
{
    if(i == 1 && j == n)
        {
           // cout <<"k"<<r[i][j]<<" is the root"<<endl;
            dep += p[r[i][j]];
        }
    if(i == j )
    {
        //cout<<"d"<<i-1<<" is the left child of k"<<i<<endl;
        dep += depth*q[i-1];
        //cout<<"d"<<i<<" is the right child of k"<<i<<endl;
        dep += depth*q[i];
    }
    else if(r[i][j] == i)
    {
        //cout<<"d"<<i-1<<" is the left child of k"<<i<<endl;
        dep += depth*q[i-1];
        //cout<<"k"<<r[i+1][j]<<" is the right child of k"<<r[i][j]<<endl;
        dep += depth*p[r[i+1][j]];
        PRINT_OPTIMAL_BST(i+1,j,depth+1);
    }
    else if(r[i][j] == j)
    {
        //cout<<"k"<<r[i][j-1]<<" is the left child of k"<<r[i][j]<<endl;
        dep += depth*p[r[i][j-1]];
        PRINT_OPTIMAL_BST(i,j-1,depth+1);
        //cout<<"d"<<j<<" is the right child of k"<<j<<endl;
        dep += depth*q[j];
    }
    else
    {
        //cout<<"k"<<r[i][r[i][j]-1]<<" is the left child of k"<<r[i][j]<<endl;
        dep += depth*p[r[i][r[i][j]-1]];
        PRINT_OPTIMAL_BST(i,r[i][j]-1,depth+1);
        //cout<<"k"<<r[r[i][j]+1][j]<<" is the right child of k"<<r[i][j]<<endl;
        dep += depth*p[r[r[i][j]+1][j]];
        PRINT_OPTIMAL_BST(r[i][j]+1,j,depth+1);
    }
}
void OPTIMAL_BST()
{
    for(int i = 1; i <= n+1; i++)
    {
        e[i][i-1] = w[i][i-1] = q[i-1];
    }
    for(int l = 1; l <= n; l++)
    {
        for(int i = 1; i <= n-l+1; i++)
        {
            int j = i+l-1;
            e[i][j] = 0x7ffffff;
            w[i][j] = w[i][j-1] + p[j]+q[j];
            for(int k = i ; k <= j ; k++)
            {
                T t = e[i][k-1]+e[k+1][j]+w[i][j];
                if(t < e[i][j])
                {
                    e[i][j] = t;
                    r[i][j] = k;
                }
            }
        }
    }
    PRINT_OPTIMAL_BST(1,n,2);
}

int main()
{
    while(cin>>n)
    {
        memset(w,0,sizeof(w));
        memset(e,0,sizeof(e));
        memset(r,0,sizeof(r));
        dep = 0;
        int sum = 0;
        for(int i = 1; i <= n; i++) cin>>p[i],sum+=p[i];
        for(int i = 0; i <= n; i++) cin >>q[i],sum+=q[i];
        //for(int i = 1; i <= n; i++) p[i]=(float)a[i]/(float)sum;
        //for(int i = 0; i <= n; i++) q[i]=(float)b[i]/(float)sum;
        OPTIMAL_BST();
        printf("%d
",(int)(dep));
    }

    return 0;
}