(PAT 1004) Counting Leaves (树的广度优先)

左手的ㄟ右手 2022-03-21 05:20 170阅读 0赞

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

### Input Specification: ###

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where `ID` is a two-digit number representing a given non-leaf node, `K` is the number of its children, followed by a sequence of two-digit `ID`'s of its children. For the sake of simplicity, let us fix the root ID to be `01`.

The input ends with N being 0. That case must NOT be processed.

### Output Specification: ###

For each test case, you are supposed to count those family members who have no child **for every seniority level** starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where `01` is the root and `02` is its only child. Hence on the root `01` level, there is `0` leaf node; and on the next level, there is `1` leaf node. Then we should output `0 1` in a line.

### Sample Input: ###

2 1
    01 1 02

### Sample Output: ###

0 1

解题思路:

先把树建起来,然后利用广度优先遍历,统计每一层中叶子结点的个数(保存到哈对应的哈希表中),最后输出哈希表

我们在广度优先遍历时可以保存层数

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <string>
    #include <string.h>
    using namespace std;
    int hashTable[1010];
    int N, M;
    int Depth = 0;
    struct FNode{
    	vector<int> child;
    }FTree[1010];
    
    void BFS_F(int root) {
    	queue<int> bfs_tree;
    	queue<int> bfs_layer;
    	bfs_layer.push(0);
    	bfs_tree.push(root);
    	while (!bfs_tree.empty()) {
    		int curLayer = bfs_layer.front();
    		int curNode = bfs_tree.front();
    		if (FTree[curNode].child.empty()) {
    			hashTable[curLayer]++;
    		}
    		Depth = curLayer;
    		bfs_tree.pop();
    		bfs_layer.pop();
    		for (auto x : FTree[curNode].child) {
    			bfs_tree.push(x);
    			bfs_layer.push(curLayer + 1);
    		}
    	}
    }
    int main() {
    	memset(hashTable, 0, sizeof(hashTable));
    	scanf("%d %d", &N, &M);
    	for (int i = 0; i < M; ++i) {
    		int ID,k;
    		scanf("%d %d", &ID,&k);
    		for (int j = 0; j < k; ++j) {
    			int chi;
    			scanf("%d", &chi);
    			FTree[ID].child.push_back(chi);
    		}
    	}
    	BFS_F(1);
    	for (int i = 0; i <= Depth; ++i) {
    		printf("%d", hashTable[i]);
    		if (i < Depth) printf(" ");
    	}
    	system("PAUSE");
    	return 0;
    }