1004. Counting Leaves (30)

pat。

这个题目主要就是存储树以及遍历树。

解题心得：

1、可能对c++不是很熟，所以一开始我就循序渐进而用了教材上所给的结构体+指针来写。不过，确实有点晕。

2、我的树存储方法是：孩子存储法，也就是把每个节点都存在一个数组，然而后面链接着他的所有儿子。

3、BFS的原理就是树的层次遍历，所以我还是用了层次遍历，中间再加了一些题目所需要的语句。

4、需要注意的地方：

①：结构体指针用之前需要为它申请内存空间。。。。

②：需要手动把链表最后一个节点置空，否则虽然vs显示无法计算内存地址，但它还是会走到哪儿，然后终止程序。

③：由于我的存储方法是孩子存储法，所以后面循环的不是tree[myParent]，而是他的第一个儿子。

struct node *firstChild;firstChild = (struct node*)malloc(sizeof(struct node));firstChild = tree[myParent].firstChild;while(NULL != firstChild){firstChild->level = level+1;//deeper and incrementmyQueue.push(firstChild);firstChild = firstChild->nextChild;}

代码如下：

#include"stdio.h"#include"string.h"#include"math.h"#include<iostream>#include<vector>#include<queue>using namespace std;#define INF 0x7FFFFFFF//the storage structure of tree,you can choose other ways to store,also.typedef struct node{int value;int level;struct node *nextChild;}*Node;struct PTree{int parent;struct node *firstChild;};vector<struct PTree> tree;vector<int> nonLeafs;queue<Node> myQueue;//level traversal,that is BFSvoid levelOrder(){int leaf = 0;struct node *parent;//get the first node and push it into the queue.parent = (struct node*)malloc(sizeof(struct node));int currentStep = 0;parent->value = tree[1].parent;parent->nextChild = tree[1].firstChild;parent->level = 0;myQueue.push(parent);while(!myQueue.empty()){   //loop if not emptystruct node *tempNode;tempNode = (struct node*)malloc(sizeof(struct node));tempNode = myQueue.front();myQueue.pop();//get the first nodeint myParent = tempNode->value;int level = tempNode->level;//modify the level and leaf numberif(level > currentStep){currentStep = level;nonLeafs.push_back(leaf);leaf = 0;}//modify key of leaf if it has no child.if(tree[myParent].firstChild == NULL){leaf++;}//get the second node and push it into the queue,not the first!struct node *firstChild;firstChild = (struct node*)malloc(sizeof(struct node));firstChild = tree[myParent].firstChild;while(NULL != firstChild){firstChild->level = level+1;//deeper and incrementmyQueue.push(firstChild);firstChild = firstChild->nextChild;}}//add the last node's leaf number.nonLeafs.push_back(leaf);}int main(){int nodes,nonLeafNodes;while(scanf("%d%d",&nodes,&nonLeafNodes) != EOF){tree.resize(nodes+1);     //size from 1~nodesfor(int i = 1;i < nonLeafNodes+1;i++){int parentNode,nodesSize;scanf("%d%d",&parentNode,&nodesSize);tree[parentNode].parent = parentNode;struct node *tempNode;tempNode = (struct node*)malloc(sizeof(struct node));tree[parentNode].firstChild = NULL;for(int j = 0;j < nodesSize;j++){// concat the tree.int tempChild;scanf("%d",&tempChild);if(j==0){tempNode->value = tempChild;tree[parentNode].firstChild = tempNode;if(j == nodesSize -1){tempNode->nextChild = NULL;}continue;}struct node *temp2Node;temp2Node = (struct node*)malloc(sizeof(struct node));temp2Node->value = tempChild;tempNode->nextChild = temp2Node;tempNode = temp2Node;if(j == nodesSize -1){tempNode->nextChild = NULL;}}}levelOrder();   //traversal//print outfor(int i = 0;i < nonLeafs.size();i++){if(i == nonLeafs.size()-1){printf("%d",nonLeafs[i]);}else{printf("%d ",nonLeafs[i]);}}}return 0;}

当然我觉得作为acm，其实可以不用用指针，网上看到一位大牛，然后他的方法并没有用链表，而是通过c++的stl库中工具完成，我觉得比我的好。

http://download.csdn.net/detail/anla_/9728778