题目解释：

解题思路：

ac代码：

多看多做，才能更加熟悉树方面的知识！Very important！！

1053 Path of Equal Weight （30 分）

题目地址：https://pintia.cn/problem-sets/994805342720868352/problems/994805424153280512

题目解释：

对于每个结点，上面是结点的编号，用二位数字表示（two-digit）,下面是该结点的权值，先寻找所有的路径，使路径上所有结点的权值之和是给定的s，并按非递减顺序输出这些路径上经过的权值。

如果对于两条路径，A1=B1,.....Ai-1=B i-1,Ai>Bi那么A路径比B路径大.

解题思路：

1）用path[i]存路径上第i个结点的编号，最后在一次输出编号对应的weight，即node[path[i]].weight

2）对于某个结点的子孩子，要先对这些子孩子的weight排序，即现遍历weight大的子孩子，最后也先输出weight大的子孩子，即实现了非递减得输出路径

3）dfs遍历树

ac代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#define maxn 110
using namespace std;
struct NODE{int weight;vector<int> child;
}node[maxn];//结点数组
bool cmp(int a,int b)//weight从大到小输出，先排序
{return node[a].weight>node[b].weight;
}
int n,m,s;
int path[maxn];//path[i]表示路径上第i个结点的编号
void dfs(int index,int numnode,int sum)//访问结点为index，当前结点个数numnode,和sum
{if(sum>s) return ;if(sum==s){if(node[index].child.size()!=0)//非叶子结点return ;for(int i=0;i<numnode;i++){printf("%d",node[path[i]].weight);if(i<numnode-1) printf(" ");else printf("\n");}return ;}for(int i=0;i<node[index].child.size();i++){int child=node[index].child[i];path[numnode]=child;dfs(child,numnode+1,sum+node[child].weight);}
}
int main()
{scanf("%d%d%d",&n,&m,&s);for(int i=0;i<n;i++)scanf("%d",&node[i].weight);int id,k,child;for(int i=0;i<m;i++){scanf("%d%d",&id,&k);for(int j=0;j<k;j++){scanf("%d",&child);node[id].child.push_back(child);//向vector中存入孩子}sort(node[id].child.begin(),node[id].child.end(),cmp);//对每一个结点的子孩子们的weight排序}path[0]=0;dfs(0,1,node[0].weight);return 0;
}

多看多做，才能更加熟悉树方面的知识！Very important！！

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PAT-A1053:Path of Equal Weight(普通树的遍历和非递减路径的输出）

题目解释：

解题思路：

ac代码：

多看多做，才能更加熟悉树方面的知识！Very important！！

PAT-A1053:Path of Equal Weight(普通树的遍历和非递减路径的输出）相关推荐

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