PAT (Advanced Level) 1053. Path of Equal Weight (30) 求树根到叶子和为所给数值的路径，DFS后排序

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 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 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor i=1, ... k, and A_k+1 > B_k+1.

Sample Input:

20 9 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

DFS后进行使用STL的sort进行排序，自定义比较方法，注意STL的sort规定相等时必须返回false，以保证稳定排序。

/*2015.7.27cyq*/#include <iostream>#include <vector>#include <algorithm>#include <fstream>using namespace std;//fstream fin("case1.txt");//#define cin finstruct TNode{int val;vector<int> next;};void dfs(const vector<TNode> &tree,int root,int target,vector<int> &path,vector<vector<int> > &result){path.push_back(tree[root].val);target-=tree[root].val;if(tree[root].next.empty()){//叶子结点if(target==0)result.push_back(path);}else{for(auto it=tree[root].next.begin();it!=tree[root].next.end();it++){dfs(tree,*it,target,path,result);}}target+=tree[root].val;path.pop_back();}bool cmp(const vector<int> &a,const vector<int> &b){if(a==b)//注意STL的sort规定相等时必须返回falsereturn false;int n=a.size();int m=b.size();int i=0;while(i<n&&i<m){if(a[i]>b[i])return true;else if(a[i]<b[i])return false;else i++;}if(i==m)return true;return false;}int main(){int N,M,S;cin>>N>>M>>S;vector<TNode> tree(N);for(int i=0;i<N;i++)cin>>tree[i].val;int addr,num,x;for(int i=0;i<M;i++){cin>>addr>>num;for(int i=0;i<num;i++){cin>>x;tree[addr].next.push_back(x);}}vector<int> path;vector<vector<int> > result;dfs(tree,0,S,path,result);sort(result.begin(),result.end(),cmp);for(auto it=result.begin();it!=result.end();++it){auto ik=(*it).begin();cout<<*ik;for(ik=(*it).begin()+1;ik!=(*it).end();ik++){cout<<" "<<*ik;}cout<<endl;}return 0;}