Acdream 1412 2-3 Trees, acdream1412

Acdream 1412 2-3 Trees, acdream1412

2-3 Trees Time Limit:12000/6000 MS (Java/Others) Memory Limit:128000/64000 KB (Java/Others) SubmitStatusProblem Description

2-3 tree is an elegant data structure partitioned ted by John Hopcroft. it is designed to implement the same functionality as the binary search tree. 2-3 tree is an ordered rooted tree with the following properties:

• The root and each internal vertex have either 2 or 3 children;
• The distance from the root to any leaf of the tree is the same.

The only exception is the tree that contains exactly one vertex-in this case the root of the tree is the only vertex, and it is simultaneously a leaf, I. e. has no children. the main idea of the described properties is that the tree with l leaves has the height O (log l ).
Given the number of leaves l there can be several valid 2-3 trees that have l leaves. For example, the picture below shows the two possible 2-3 trees with exactly 6 leaves.

Given l find the number of different 2-3 trees that have l leaves. Since this number can be quite large, output it modulo r.

Input file contains two integer numbers: l and r (1 ≤ l ≤ 5 000, 1 ≤ r ≤109 ). output one number-the number of different 2-3 trees with exactly l leaves modulo r. sample Input

6 10000000007 1000000000

Sample Output

23

Question and code:

This is a DP question. In fact, it is to find the solution of 2 x + 3 * y = n on each layer, then, push the layer to the top layer 2 * x' + 2 * y' = x + y, and so on until the top layer.

Note: The numbers 2 2 3 and 2 3 2 are different, so the number of combinations is used. I started to open the long array directly, burst the memory, and then changed it to int, and then the array was opened as small as 1/8.

I also looked at other people's code and used the scrolling array directly to save more memory and speed.

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;typedef long long LL;LL sum[5010];int bt[2510][1300];int N,R;bool flag[5010];struct node{    int x,y;    node(int X,int Y)    {        x=X;y=Y;    }};struct Root{    vector<node> nt;}r[5010];void get(int n){    if(flag[n]) return;    int j;    for(int i=0;i*2<=n;i++)    {        j=n-i*2;        if(j>=0&&j%3==0)        r[n].nt.push_back(node(i,j/3));    }    flag[n]=true;}void init(){    memset(sum,-1,sizeof(sum));    memset(flag,false,sizeof(flag));    sum[1]=0;sum[2]=1;sum[3]=1;    bt[1][0]=bt[1][1]=1;bt[1][2]=0;    int bN=N/2+1;    int bM=bN/2+1;    get(N);    for(int i=2;i<=bN;i++)    {        bt[i][0]=1;        for(int j=1;j<=bM;j++)        {            bt[i][j]=(bt[i-1][j-1]+bt[i-1][j])%R;        }    }}LL dfs(int n){    if(sum[n]>=0) return sum[n];    int len=r[n].nt.size();    LL ans=0;int k;    for(int i=0;i<len;i++)    {        k=r[n].nt[i].x+r[n].nt[i].y;        get(k);        ans=(ans+bt[k][min(r[n].nt[i].x,r[n].nt[i].y)]*dfs(k))%R;    }    //printf("%d  %d\n",n,ans);    sum[n]=ans;    return ans;}int main(){    while(scanf("%d%d",&N,&R)!=EOF)    {        if(N==1)        {            printf("%d\n",N%R);            continue;        }        init();        LL an=dfs(N);        printf("%lld\n",an);    }    return 0;}