Question 1520: Tree sub-structure-9 degrees

Description:

Enter Two Binary Trees A and B to determine whether B is a sub-structure.

Input:

The input may contain multiple test examples. The input ends with EOF.
For each test case, the input first line is an integer n, m (1 <= n <= 1000, 1 <= m <= ): N indicates the number of nodes of the Binary Tree A to be input (the number of nodes starts from 1), and M indicates the number of nodes of the Binary Tree B to be input (the number of nodes starts from 1 ). The next row contains N numbers, each representing the I-th element of the tree, and the next n rows. The first Ki number represents the number of children on the I-th node, next there are ki trees, representing the node I child node number. The next m + 1 line is the same as the description of tree.

Output:

Corresponding to each test case,
If B is a, the subtree outputs "yes" (excluding quotation marks ). Otherwise, "no" is output (excluding quotation marks ).

Sample input:

7 38 8 7 9 2 4 72 2 32 4 5002 6 7008 9 22 2 3001 12030

Sample output:

YESNO

Tip:

B is not a subtree of any tree.

Recommendation index :※

Source: http://ac.jobdu.com/problem.php? PID = 1, 1520

The algorithm itself is difficult. Pay attention to the following points:

1. Be careful when using the vector and time out ..

2. The default subtree condition of the question is: None, left and right. Don't think it's complicated.

#include<iostream>#include<string.h>#include <stdio.h>#include<stdlib.h>using namespace std;const int N=1001;int a_edge[N][2],b_edge[N][2];int a_val[N],b_val[N];int n,m;bool dfs_sub(int a,int b){//sub tree is fitif(b==0)        return true;if(a_val[a]!=b_val[b]||false==dfs_sub(a_edge[a][0],b_edge[b][0])||false==dfs_sub(a_edge[a][1],b_edge[b][1]))return false;return true;}bool is_sub(){    int i;for(i=1;i<=n;i++){if(a_val[i]==b_val[1]){if(true==dfs_sub(i,1))return true;}}return false;}int main(){    while(scanf("%d%d",&n,&m)!=EOF){memset(a_val,0,sizeof(a_val));memset(b_val,0,sizeof(b_val));        int i,j,num;for(i=1;i<=n;i++)scanf("%d",&a_val[i]);for(i=1;i<=n;i++){scanf("%d",&num);for(j=0;j<num;j++)scanf("%d",&a_edge[i][j]);}for(i=1;i<=m;i++)scanf("%d",&b_val[i]);for(i=1;i<=m;i++){int num,tmp;scanf("%d",&num);for(j=0;j<num;j++)scanf("%d",&b_edge[i][j]);}if(m!=0&&n!=0&&true==is_sub())printf("YES\n");elseprintf("NO\n");    }    return 0;}

In the version with the last case timeout of a vector attached:

#include<iostream>#include<vector>#include<string>#include <stdio.h>#include<stdlib.h>using namespace std;vector<vector<int> > a_edge;vector<vector<int> > b_edge;int *a_val,*b_val;int n,m;bool dfs_sub(int a_node,int b_node){//sub tree is fit    if(a_val[a_node]==b_val[b_node]){if(b_edge[b_node].size()<a_edge[a_node].size()){int i,j;if(b_edge[b_node].size()==0)                return true;else{                if(dfs_sub(a_edge[a_node][0],b_edge[b_node][0])==true)return true;            }        }else if(b_edge[b_node].size()==a_edge[a_node].size()){int i;bool flag=false;for(i=0;i<b_edge[b_node].size();i++){if(false==dfs_sub(a_edge[a_node][i],b_edge[b_node][i]))return false;}return true;}    }    return false;}bool is_sub(){    int i;for(i=1;i<=n;i++){if(a_val[i]==b_val[1]){if(true==dfs_sub(i,1))return true;}}return false;}int main(){    while(scanf("%d%d",&n,&m)!=EOF){        a_val=new int [n+1];        b_val=new int [m+1];        a_edge.resize(n+1);        b_edge.resize(m+1);        int i,j;for(i=1;i<=n;i++)scanf("%d",&a_val[i]);for(i=1;i<=n;i++){int num,tmp;scanf("%d",&num);for(j=1;j<=num;j++){scanf("%d",&tmp);a_edge[i].push_back(tmp);}}for(i=1;i<=m;i++)scanf("%d",&b_val[i]);for(i=1;i<=m;i++){int num,tmp;scanf("%d",&num);for(j=1;j<=num;j++){scanf("%d",&tmp);b_edge[i].push_back(tmp);}}if(m!=0&&n!=0&&true==is_sub())printf("YES\n");elseprintf("NO\n");    }    return 0;}