1053 Path of Equal Weight

1. 以下两组关系很大的概念

树的深度优先搜索 - 先根遍历 - 递归

树的广度优先搜索 - 层序遍历 - 非递归

本题考察的是前者，我设置了这样一个结构体

struct Prestruct{int totalWei = 0;vector<int> pre;
};Prestruct pre[maxn];

pre[idx].pre向量存放父节点到当前结点的权重

递归式：父节点的pre向量赋给子节点

递归边界：当前结点为叶子结点，比对该结点的权重和与目标是否相等，如果相等，将其加入

vector<vector<int> > ans;

注意到第一次我写这道题还用数组存储，所以最后的非升序输就处理得又臭又长，而vector<int>自身是可以比较的。

AC代码

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<bits/stdc++.h>
#include<stdlib.h>
#include<time.h>
#include<vector>
#include<set>
#include<string>
#include<tr1/unordered_map>using namespace std;
using namespace std::tr1;
typedef long long LL;const int maxn = 107;
const int MOD = 1000000007;
const int INF = 1000000000;//INF:下确界
const LL SUP = (1LL<<63)-1;//SUP:上确界
const double eps = 1e-5;struct Node{int v;vector<int> children;
};Node node[maxn];int target;struct Prestruct{int totalWei = 0;vector<int> pre;
};Prestruct pre[maxn];vector<vector<int> > ans;bool cmp(vector<int> a,vector<int> b){return a>b;
}void DFS(int root){pre[root].pre.push_back(node[root].v);pre[root].totalWei += node[root].v;if(node[root].children.size()==0){if(pre[root].totalWei==target){ans.push_back(pre[root].pre);}}for(int i=node[root].children.size()-1;i>=0;i--){int u = node[root].children[i];pre[u].pre = pre[root].pre;pre[u].totalWei = pre[root].totalWei;DFS(u);}
}int main(){int vNum,eNum;cin>>vNum>>eNum>>target;for(int i=0;i<vNum;i++){cin>>node[i].v;}while(eNum--){int v,num,vc;cin>>v>>num;while(num--){cin>>vc;node[v].children.push_back(vc);}}DFS(0);//    for(int i=0;i<vNum;i++){
//      printf("%d\n",pre[i].totalWei);
//  }sort(ans.begin(),ans.end(),cmp);for(int i=0;i<ans.size();i++){for(int j=0;j<ans[i].size();j++){printf("%d%s",ans[i][j],j!=ans[i].size()-1?" ":"\n");}}return 0;
}

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