1053 Path of Equal Weight(超级无敌详细注释+45行代码)

分数 30

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作者 CHEN, Yue

单位浙江大学

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L.

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in the following figure: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in the figure.

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, M (<N), the number of non-leaf nodes, and 0<S<230, the given weight number. The next line contains N positive numbers where Wi (<1000) corresponds to the tree node Ti. 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 00.

Output Specification:

For each test case, print all the paths with weight S in non-increasing order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end of the line.

Note: sequence {A1,A2,⋯,An} is said to be greater than sequence {B1,B2,⋯,Bm} if there exists 1≤k<min{n,m} such that Ai=Bi for i=1,⋯,k, and Ak+1>Bk+1.

Sample Input:

20 9 24
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
00 4 01 02 03 04
02 1 05
04 2 06 07
03 3 11 12 13
06 1 09
07 2 08 10
16 1 15
13 3 14 16 17
17 2 18 19

Sample Output:

Special thanks to Zhang Yuan and Yang Han for their contribution to the judge's data.

代码长度限制

16 KB

时间限制

400 ms

内存限制

64 MB

#include<bits/stdc++.h>
using namespace std;
const int N=105;
int n,m,s;
int w[N];//存放结点权值
vector<int>v[N];//存放各非叶节点的孩子节点
vector<vector<int>>ans;//存放路径
void dfs(int u,int sum,vector<int>&path){//深度遍历
   bool leaf=true;
   if(v[u].size())leaf=false;//若有孩子则不是叶结点
   if(leaf){//如果是叶结点
       if(sum==s)ans.push_back(path);//若此时权值符合，则放到ans里
   }
   else{//若不是叶结点
           if(v[u].size()){//若该节点有孩子
               for(int j=0;j<v[u].size();j++){//暴搜
                   path.push_back(w[v[u][j]]);//把当前结点的权值加入路径
                   dfs(v[u][j],sum+w[v[u][j]],path);//深度遍历
                   path.pop_back();//恢复现场
           }
       }
   }
}
int main(){
   cin>>n>>m>>s;
   for(int i=0;i<n;i++)cin>>w[i];//输入各结点权值
   while(m--){//输入m个非叶节点
       int id,k;
       cin>>id>>k;
       while(k--){//保存结点的孩子
           int kid;
           cin>>kid;
           v[id].push_back(kid);
       }
   }
   vector<int>path({w[0]});//权值路径，一开始只有根结点权值
   dfs(0,w[0],path);//深度遍历
   sort(ans.begin(),ans.end(),greater<vector<int>>());//按字典序从大到小排列
   for(auto t:ans){//迭代器遍历
       cout<<t[0];
       for(int i=1;i<t.size();i++)cout<<" "<<t[i];
       cout<<endl;
   }
   return 0;
}