PAT-A-刷题记录

刷刷刷,要有多大的成就就得有多大的努力啊。

DFS 深度搜索

什么是深度搜索尼?看下面的图,然后我进行一个搜索。

  1. 先搜索的是1234然后发现4,会回到3寻找别的路径
  2. 发现5这个路径进入,然后发现5也已经是死胡同
  3. 发现3已经结束没有别的路径在往上2,有一个路径是6,然后6也是死胡同,然后再往上,2也已经结束
  4. 在往上到1,1的另一个路径为7,7也为死胡同了
  5. 在到另一个路径8,接下来就是一样了就不说了。

那么这个算法的实现是怎么样的尼。

  1. 先是递归,为什么用递归,作为算法菜鸡只知道这样很好用,然后递归实现栈操作很容易。
  2. 边界条件是什么?在这里面边界条件就是下一个子结点为空。

这里以PAT-A-1004为例:

PAT-A-1004

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:

1
2
2 1
01 1 02

Sample Output:

1
0 1

题目好晦涩难懂的英语,渣渣看了好久。。

思路上,借鉴了柳神的思路,先是利用vector创建二元数组然后利用dfs进行一个搜索,去搜索叶节点。而边界条件是什么?边界条件就是当vector.size()等于0的时候,这个时候说明这个index后面没有东东了,所以就可以确定这是叶结点。

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#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

vector<int> v[100];

int book[100], maxdepth = -1;
void dfs(int index,int depth){
if(v[index].size() == 0){
book[depth]++;
maxdepth = max(maxdepth,depth);
return ;
}
for(int i = 0;i<v[index].size();i++){
dfs(v[index][i],depth+1);
}
}

int main(){
int n,m,k,note,c;
scanf("%d %d",&n,&m);
for(int i = 0;i < m;i++){
scanf("%d %d",&note,&k);
for(int j = 0; j < k;j++){
scanf("%d",&c);
v[note].push_back(c);
}
}
dfs(1,0);
printf("%d",book[0]);
for(int i = 1;i<=maxdepth ;i++){
printf(" %d",book[i]);
}
}

贴上代码,说起来其实dfs本质应该是,递归运算向子节点不断探索,到了边界条件再返回。具体实现可以利用vector去创建二维数组来实现。