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 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??,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

總結(jié)：

用靜態(tài)樹的構(gòu)造邏輯上還是比較清晰的，就是一個數(shù)組，然后數(shù)組下有若干個子結(jié)點，通過子結(jié)點的數(shù)值找到數(shù)組中下標(biāo)

這里的樹和二叉樹的不同之處在于，二叉樹只有兩個子結(jié)點，而樹的子結(jié)點是不確定的，基于此，在構(gòu)造結(jié)構(gòu)的時候把子結(jié)點構(gòu)造為用vector容器來存儲，并用push_back和pop_back來插入刪除

一般來說，輸出的時候是要排序的，可能按結(jié)點的權(quán)重從大到小或者從小到大，本題給出一種思路，就是在輸入的時候就把子結(jié)點按權(quán)重按要求排好序，這樣在輸出的時候就可以直接輸出了。

在DFS中，首先要設(shè)置邊界

一般來說，要有一個記錄路徑多少的數(shù)，這是為了輸出方便，因為數(shù)與數(shù)之間需要用空格隔開，但是最后一個數(shù)后面不許出現(xiàn)空格。這就要求要做一個判別，這里用sumNode來判別的

DFS是一個遞歸的結(jié)構(gòu)

代碼：

#include <stdio.h> #include <vector> #include <algorithm> using namespace std; const int maxn=110; int total,row,sum; int path[maxn]; //路徑 struct Node{int data;vector<int> child; }node[maxn];bool cmp(int a,int b){ //提前排序，有助于按降序輸出 return node[a].data>node[b].data; } void DFS(int index,int sumNode,int s){ //結(jié)點，路徑中的個數(shù)，和 if(s>sum) return;if(s==sum){if(node[index].child.size()!=0) return;for(int i=0;i<sumNode;i++){printf("%d",node[path[i]].data);if(i<sumNode-1) printf(" ");else printf("\n");}return;}for(int i=0;i<node[index].child.size();i++){int child=node[index].child[i];path[sumNode]=child;DFS(child,sumNode+1,s+node[child].data);} } int main(){scanf("%d%d%d",&total,&row,&sum);for(int i=0;i<total;i++){scanf("%d",&node[i].data);}int id,num,child;for(int i=0;i<row;i++){scanf("%d %d",&id,&num);for(int j=0;j<num;j++){scanf("%d",&child);node[id].child.push_back(child);}sort(node[id].child.begin(),node[id].child.end(),cmp); //先對整個子樹按權(quán)重從左到右由大向小排序，那之后的輸出就不必再作比較了 }path[0]=0;//存放結(jié)點DFS(0,1,node[0].data); return 0; }

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總結(jié)

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