Given a non-empty tree with root R, and with weight Wiassigned to each tree node Ti. Theweight 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 innon-increasingorder. 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 begreater thansequence {B1, B2, …, Bm} if there exists 1 <= k < min{n, m} such that Ai= Bifor i=1, … k, and Ak+1> Bk+1.
#include <stdio.h>#include <stdlib.h>#include <vector>#include <algorithm>using namespace std;#define N 100typedef struct {int data;vector<int> child;}Node;Node node[N];int n,m,s;int ans[N];void DFS (int a,int num,int sum);bool cmp (int a,int b);int main (){int i,temp;scanf("%d %d %d",&n,&m,&s);for( i=0;i<n;i++){scanf("%d",&temp);node[i].data=temp;}int id,num,childs,j;for( i=0;i<m;i++){scanf("%d %d",&id,&num);for( j=0;j<num;j++){scanf("%d",&childs);node[id].child.push_back(childs);}sort(node[id].child.begin(),node[id].child.end(),cmp);}ans[0]=node[0].data;DFS(0,1,node[0].data);system("pause");return 0;}bool cmp (int a,int b){return node[a].data>node[b].data;}void DFS (int a,int num,int sum){int i;if( sum>s) return;if( sum==s){if( node[a].child.size()==0){for( i=0;i<num-1;i++) printf("%d ",ans[i]);printf("%d\n",ans[i]);}return;}for( i=0;i<node[a].child.size();i++){int child=node[a].child[i];ans[num]=node[child].data;DFS(child,num+1,sum+node[child].data);}}
，勇于接受自己的失败，告诉自己，这就是自己的现实，
