ZOJ - 3602 Count the Trees 二叉树同构

A binary tree is a tree data structure in which each node has at most two child nodes, usually distinguished as "left" and "right". A subtree of a tree T is a tree consisting of a node in T and all of its descendants in T. Two binary trees are called identical if their left subtrees are the same(or both having no left subtree) and their right subtrees are the same(or both having no right subtrees).

According to a recent research, some people in the world are interested in counting the number of identical subtree pairs, each from the given trees respectively.

Now, you are given two trees. Write a program to help to count the number of identical subtree pairs, such that the first one comes from the first tree and the second one comes from the second tree.

Input

There are multiple test cases. The first line contains a positive integer T (T ≤ 20) indicating the number of test cases. Then T test cases follow.

In each test case, There are two integers n and m (1 ≤ n, m ≤ 100000) indicating the number of nodes in the given two trees. The following n lines describe the first tree. The i-th line contains two integers u and v (1 ≤ u ≤ n or u = -1, 1 ≤ v ≤ n or v = -1) indicating the indices of the left and right children of node i. If uor v equals to -1, it means that node i don't have the corresponding left or right child. Then followed by m lines describing the second tree in the same format. The roots of both trees are node 1.

Output

For each test case, print a line containing the result.

Sample Input

2
2 2
-1 2
-1 -1
2 -1
-1 -1
5 5
2 3
4 5
-1 -1
-1 -1
-1 -1
2 3
4 5
-1 -1
-1 -1
-1 -1

Sample Output

1
11

       求a,b树中相同结构的二叉子树对数的数量。

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int a[maxn][2], b[maxn][2];
int sum[maxn]; long long ans;
map<pair<int, int>, int>G;
int cnt = 1;
pair<int, int> makepair(int a, int b)
{
	pair<int, int> p;
	p.first = a;
	p.second = b;
	return p;
}

int dfs_a(int u) {
	int son1, son2;
	if (a[u][0] != -1)
		son1 = dfs_a(a[u][0]);
	else son1 = -1;

	if (a[u][1] != -1)
		son2 = dfs_a(a[u][1]);
	else son2 = -1;
	if (G.find(makepair(son1, son2)) == G.end()) {
		G[makepair(son1, son2)] = cnt++;
	}
	int id = G[makepair(son1, son2)];
	sum[id]++;
	return id;
}

int dfs_b(int u) {
	int son1, son2;
	if (b[u][0] != -1)
		son1 = dfs_b(b[u][0]);
	else son1 = -1;
	if (b[u][1] != -1)
		son2 = dfs_b(b[u][1]);
	else son2 = -1;
	if (G.find(makepair(son1, son2)) == G.end()) {
		G[makepair(son1, son2)] = 0;
	}
	int id = G[makepair(son1, son2)];
	ans += sum[id];
	return id;
}

int main() {
	int te;
	scanf("%d", &te);
	while (te--) {
		ans = 0;
		int n, m; cnt = 1; G.clear();
		scanf("%d%d", &n, &m);
		for (int i = 0; i <= max(n, m); i++)sum[i] = 0;
		for (int i = 1; i <= n; i++) {
			scanf("%d%d", &a[i][0], &a[i][1]);
		}
		for (int i = 1; i <= m ; i++) {
			scanf("%d%d", &b[i][0], &b[i][1]);
		}
		
		dfs_a(1);
		dfs_b(1);
		printf("%lld\n", ans);
	}
	return 0;
}