noi 2009 二叉查找树 动态规划

思路：

先把权值离散化

按数据值排序

sum[i]为前i个节点频度和

dp[i][j][w]表示把节点[i,j]合并成一颗根节点权值不小于w的子树所需的访问代价与修改代价的最小和

dp[i][j][w]=min(dp[i][k-1][w]+dp[k+1][j][w]+sum[j]-sum[i-1]+K,dp[i][k-1][a[k].weight]+dp[k+1][j][a[k].weight]+sum[j]-sum[i-1](a[k].weight>=w))

(i<=k<=j)

ans=dp[1][n][1];

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
#define MAXN 80
#define INF 987654321
int n,m;
struct node
{
    int key,weight,frequence;
}a[MAXN];
int sum[MAXN];
int dp[MAXN][MAXN][MAXN];
bool cmp1(const node &A,const node &B)
{
    return A.weight<B.weight;
}
bool cmp2(const node &A,const node &B)
{
    return A.key<B.key;
}
int main()
{
    memset(dp,0,sizeof(dp));
    scanf("%d%d",&n,&m);
    int i,j,k,w,t;
    for(i=1;i<=n;i++)
        scanf("%d",&a[i].key);
    for(i=1;i<=n;i++)
        scanf("%d",&a[i].weight);
    for(i=1;i<=n;i++)
        scanf("%d",&a[i].frequence);
    sort(a+1,a+n+1,cmp1);
    for(i=1;i<=n;i++)
        a[i].weight=i;
    sort(a+1,a+n+1,cmp2);
    sum[0]=0;
    for(i=1;i<=n;i++)
        sum[i]=sum[i-1]+a[i].frequence;
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
        {
            if(a[i].weight>=j)
                dp[i][i][j]=a[i].frequence;
            else
                dp[i][i][j]=a[i].frequence+m;
        }
    for(w=n;w>=1;w--)
    for(t=1;t<n;t++)
    for(i=1;i+t<=n;i++)
    {
        j=i+t;
        int s=INF;
        for(k=i;k<=j;k++)
        {
                s=min(s,dp[i][k-1][w]+dp[k+1][j][w]+sum[j]-sum[i-1]+m);
                if(a[k].weight>=w)
                    s=min(s,dp[i][k-1][a[k].weight]+dp[k+1][j][a[k].weight]+sum[j]-sum[i-1]);
        }
        dp[i][j][w]=s;
    }
    int s=INF;
    printf("%d\n",dp[1][n][1]);
    return 0;
}