| 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: where 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:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2题目大意
给你树的路径和权值，请找出找从根结点到叶⼦结点路径上的权值相加之和等于Key的路径，并且从⼤到⼩输出路径
思路
DFS暴力遍历即可
C/C++
#include<bits/stdc++.h>
using namespace std;
void DFS(int now,int sum);
vector<vector<int>> result;
vector<int> child[10001],op;
int power[10001],N,M,key,a,b,c;
bool cmp(vector<int>& x,vector<int>& y){
for(int z=0;z<min(x.size(),y.size());z++){
if(power[x[z]]!=power[y[z]]) return power[x[z]] > power[y[z]];
}
return false;
}
int main()
{
cin >> N >> M >> key;
for(int z=0;z<N;z++) cin >> power[z];
while (M--){
cin >> a >> b;
while (b--){
cin >> c;
child[a].push_back(c);
}
}
DFS(0,power[0]);
if(result.size()>1) sort(result.begin(),result.end(),cmp);
for(const auto& x:result){
cout << power[0];
for(int y:x) cout << " " << power[y];
putchar('\n');
}
return 0;
}
void DFS(int now,int sum)
{
if(child[now].empty()){
if(sum==key) result.push_back(op);
return;
}else{
for(int x:child[now]) {
op.push_back(x);
DFS(x,sum+power[x]);
op.pop_back();
}
}
}
