1053 Path of Equal Weight (30 分) dfs，树

Given a non-empty tree with root R, and with weight W​i 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<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,A2,⋯,A​n} is said to be greater than sequence {B1,B2,⋯,B​m} 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

#include<iostream>
#include<cstdio>
#include<stack> 
#include<queue>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<cctype>
#include<cstdlib>
#include<ctime>
#include<unordered_map>

using namespace std;

const int MAXN = 110;

struct node {
	int weight;
	vector<int> child;
} Node[MAXN];

bool cmp(int a,int b) {
	return Node[a].weight > Node[b].weight;
}

int n,m,s;
int path[MAXN];

void DFS(int index,int numNode,int sum) {
	if(sum > s) {
		return;
	} 
	if(sum == s) {
		if(Node[index].child.size() != NULL) {
			return;
		} else {
			for(int i = 0;i < numNode;i++) {
				printf("%d",Node[path[i]].weight);
				if(i < numNode - 1) {
					printf(" ");
				} else {
					printf("\n");
				}
			}
		}
	}
	if(sum < s) {
		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);
		}
		sort(Node[id].child.begin(),Node[id].child.end(),cmp);
	}
	path[0] = 0;
	DFS(0,1,Node[0].weight);
	return 0;
}