1053. Path of Equal Weight (30)
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 Lis 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 Figure 1: 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 Figure 1.
Figure 1
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 = Bifor 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 19Sample Output:
10 5 2 7 10 4 10 10 3 3 6 2 10 3 3 6 2
深度遍历dfs，dfs函数分两部分，第一部分写达到某些条件进行返回，第二部分写遍历下一个节点。
这道题在顺序输出上有两个思路，一个是在children中添加子节点时，进行排序，使得在遍历过程中实现顺序遍历；另一个是将遍历后符合条件的结果放入一个数组中，最后在对数组排序，但是第二种方法，我在PAT上运行程序，有一个case显示段错误，在牛客网上却是正确的，我很诧异，可能是vector数组sort的问题？我不确定。
#include <bits/stdc++.h> using namespace std; int N, M, K; struct Node { int weight; vector<int> children; }nodes[102]; bool cmp(int a, int b) { return nodes[a].weight > nodes[b].weight; } bool cmp2(vector<int> vec1, vector<int>vec2) { int len1 = vec1.size(); int len2 = vec2.size(); int len = min(len1, len2); for(int i = 0; i < len; i++) { if(nodes[vec1[i]].weight < nodes[vec2[i]].weight) { return false; } else if(nodes[vec1[i]].weight > nodes[vec2[i]].weight) { return true; } } return true; } vector<int> nodesVec; //vector<int> nodesVecs[102]; //int INDEX = 0; void dfs(int index, int weights) { //cout<< "index:"<< index<< endl; if(weights > K) { return; } if(nodes[index].children.size() == 0) { if(weights == K) { //nodesVecs[INDEX] = nodesVec; //INDEX++; for(int i = 0; i < nodesVec.size(); i++) { if(i != 0) cout<< " "; cout<< nodes[nodesVec[i]].weight; } cout<< endl; } return; } for(int i = 0; i < nodes[index].children.size(); i++) { nodesVec.push_back(nodes[index].children[i]); dfs(nodes[index].children[i], weights+nodes[nodes[index].children[i]].weight); nodesVec.erase(nodesVec.end()-1); } } int main() { cin>>N>>M>>K; for(int i = 0; i < N; i++) { cin>> nodes[i].weight; } int index, s, c; for(int i = 0; i < M; i++) { cin>> index>> s; for(int j = 0; j < s; j++) { cin>> c; nodes[index].children.push_back(c); } sort(nodes[index].children.begin(), nodes[index].children.end(), cmp); } nodesVec.push_back(0); dfs(0, nodes[0].weight); /* sort(nodesVecs, nodesVecs+INDEX, cmp2); for(int i = 0; i < INDEX; i++) { for(int j = 0; j < nodesVecs[i].size(); j++) { if(j != 0) cout<< " "; cout<< nodes[nodesVecs[i][j]].weight; } cout<< endl; }*/ return 0; }