Zoj 3602 count the homogeneity (hash) of the trees tree)

Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 4705

Question: Give Two Binary Trees A and B, and ask how many pairs have the same structure as the sub-tree a in a and the sub-tree B in B.

Idea: hash. Different numbers correspond to different structures. For example, 1 represents a separate leaf node, 2 indicates that the Left subtree is a leaf node without the right subtree... every time a new subtree occurs, it is recorded, the record method is to use the DFS backtracking process to determine whether the left subtree and the right subtree have appeared (the pair is used to record the subtree, that is, the left and right subtree, two maps. The first map indicates the paing between the pair of the subtree and its labels. The second map indicates the number of occurrences of each label ).

Code:

#include <algorithm>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <ctime>#include <ctype.h>#include <iostream>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>#define eps 1e-8#define INF 0x7fffffff#define maxn 100005#define PI acos(-1.0)#define seed 31//131,1313typedef long long LL;typedef unsigned long long ULL;using namespace std;int top=0;map < pair < int , int > , int > M;map < int , LL > N ;int tree_a[maxn][2],tree_b[maxn][2],a[maxn],b[maxn];LL ans;void dfs_a(int u){    pair < int , int > p;    int left,right;    left=tree_a[u][0],right=tree_a[u][1];    if(left!=-1)        dfs_a(left);    if(right!=-1)        dfs_a(right);    if(left==-1)        p.first=-1;    else p.first=a[left];    if(right==-1)        p.second=-1;    else p.second=a[right];    if(M.find(p)==M.end())    {        M[p]=top;        a[u]=top;        N[top]=1;        top+=3;    }    else    {        a[u]=M[p];        N[M[p]]++;    }}void dfs_b(int u){    pair <int ,int > p ;    int left=tree_b[u][0],right=tree_b[u][1];    if(left!=-1)        dfs_b(left);    if(right!=-1)        dfs_b(right);    if(left==-1)        p.first=-1;    else p.first=b[left];    if(right==-1)        p.second=-1;    else p.second=b[right];    if(M.find(p)!=M.end())    {        ans+=N[M[p]];        b[u]=M[p];    }}void init(){    top=0;    ans=0;    N.clear();    M.clear();    for(int i=0;i<=100000;i++)        a[i]=-2,b[i]=-2;}int main(){    int T,n,m;    scanf("%d",&T);    while(T--)    {        init();        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)            scanf("%d%d",&tree_a[i][0],&tree_a[i][1]);        for(int i=1; i<=m; i++)            scanf("%d%d",&tree_b[i][0],&tree_b[i][1]);        dfs_a(1);        dfs_b(1);        printf("%lld\n",ans);    }    return 0;}

Zoj 3602 count the homogeneity (hash) of the trees tree)