1053 Path of Equal Weight

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本文探讨了在一个加权树中寻找所有从根节点到叶节点，且路径上节点权重之和等于给定数值的问题。通过使用广度优先搜索(BFS)，文章提供了一种有效的解决方案，并附带了代码实现。此外，还介绍了如何对找到的路径进行排序。

1053 Path of Equal Weight
题目链接
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<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.

在这里插入图片描述
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
题目大意：现有一颗树，每个结点都有一个权重，求一条从根结点到叶子结点的路径，使路径上的结点的权重之和等于给定值。求出满足条件的所有路径。
分析：求路径，可使用bfs。本题给出两种参考方法。
路径排序的话，使用sort函数。
参考代码：

#include<iostream>
#include<vector>
#include<iostream>
#include<algorithm>
using namespace std;
struct  node{
	int w;
	vector<int>child;
};
int S;
vector<int>path, temp_path;
vector<node>tree;
vector<vector<int>>paths;
void bfs(int id, int sum, int S) {
	sum += tree[id].w;
	path.push_back(tree[id].w);
	if (sum < S) {
		for (int i = 0; i < tree[id].child.size(); i++) {
			int vec = tree[id].child[i];
			bfs(vec, sum, S);
		}
		if (!path.empty()) {
			path.pop_back();
		}
		//return;
	}
	else if (sum > S) {
		path.pop_back();
		//return;
	}
	else if (sum == S) {
		if (tree[id].child.size() == 0) {
			paths.push_back(path);
			path.pop_back();
			//return;
		}
		else {
			path.pop_back();
			//return;
		}
	}
}
bool cmp(vector<int>a, vector<int>b) {
	int la = a.size(), lb = b.size();
	for (int i = 0; i < la&&i < lb; i++) {
		if (a[i] != b[i])
			return a[i] > b[i];
	}
	if (la > lb)return true;
	else return false;
}
void BFS(int id, int vecNum,int sum) {
	temp_path[vecNum] = tree[id].w;
	sum += tree[id].w;
	if (sum > S)return;
	else if (sum == S) {
		if (tree[id].child.size() == 0) {
			vector<int>ans;
			for (int i = 0; i <= vecNum; i++) {
				ans.push_back(temp_path[i]);
			}
			paths.push_back(ans);
		}
		else {  return;	}
	}
	else if (sum < S) {
		for (int i = 0; i < tree[id].child.size(); i++) {
			BFS(tree[id].child[i], vecNum + 1, sum);
		}
	}
}

int main() {
	int N, M, id, k, vec;
	cin >> N >> M >> S;
	tree.resize(N);
	temp_path.resize(N);
	for (int i = 0; i < N; i++) {
		cin >> tree[i].w;
	}
	for (int i = 1; i <= M; i++) {
		cin >> id >> k;
		for (int j = 1; j <= k; j++) {
			cin >> vec;
			tree[id].child.push_back(vec);
		}
	}
	//bfs(0, 0, S);
	BFS(0, 0, 0);
	sort(paths.begin(), paths.end(), cmp);

	for (int i = 0; i < paths.size(); i++) {
		for (int j = 0; j < paths[i].size(); j++) {
			cout << paths[i][j];
			if (j == paths[i].size() - 1)cout << endl;
			else cout << " ";
		}
	}

	return 0;
}

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