1053 Path of Equal Weight (30分)

1053 Path of Equal Weight (30分)

Given a non-empty tree with root R, and with weight W​i​​ assigned to each tree node T​i​​. 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<2​30​​, the given weight number. The next line contains N positive numbers where W​i​​ (<1000) corresponds to the tree node T​i​​. 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 {A​1​​,A​2​​,⋯,A​n​​} is said to be greater than sequence {B​1​​,B​2​​,⋯,B​m​​} if there exists 1≤k<min{n,m} such that A​i​​=B​i​​ for i=1,⋯,k, and A​k+1​​>B​k+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

#include <iostream>
#include <algorithm>
#include<vector>
#include<map>
#include<string>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<set>
#include<unordered_map>
#include<queue>
#include<climits>
#include<stack>
using namespace std;
const int maxn= 10000+10;

struct node{
	int w;	
	vector<int>child;
};

vector<node>v;
vector<int>path;

bool cmp(int a,int b){
	
	return v[a].w>v[b].w;	
}
	int n,m,s;
void dfs(int index,int num,int sum){
	if(sum>s) return ;
	if(sum==s){
		if(v[index].child.size()!=0) return ;
		for(int i=0;i<num;i++){
		
			cout<<v[path[i]].w;	
				if(i!=num-1) cout<<" ";
			else cout<<endl;
		}
		return ;
		
	}
	for(int i=0;i<v[index].child.size();i++){
		int son=v[index].child[i];
		path[num]=son;
		dfs(son,num+1,sum+v[son].w);	
	}
	
	
	
}

int main(){

	cin>>n>>m>>s;
	v.resize(n),path.resize(n);
	for(int i=0;i<n;i++) cin>>v[i].w;
	for(int i=0;i<m;i++) {
		int index;
		cin>>index;
		int k;	
		cin>>k;
		v[index].child.resize(k);
		for(int j=0;j<k;j++){
			cin>>v[index].child[j];
		}
		sort(v[index].child.begin(),v[index].child.end(),cmp);
	}

	dfs(0,1,v[0].w);	

	return 0;