1053. Path of Equal Weight (30)

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 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.

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 = Bi for 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 19
Sample Output:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

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结题思路：

根据输入，构建二叉树。
题目要求以非递增的方式输出所有路径，考虑到如果保存路径再输出，最后还要排序。就直接在建树阶段对father节点的孩子根据对应的weight进行了排序，如此，程序查找路径时的顺序就是满足条件的。
建树完成后，后序遍历输出即可。

#include<iostream>#include<cstdlib>#include<cstring>#include<string>#include<cstdlib>#include<stack> #include<algorithm> using namespace std;/*只有当child为-1时为叶子节点postOrder递归时并未对传入的roor作出区分，故将child默认为-1，bro默认为-2在树的构建阶段，就对每个非叶子节点的孩子进行了排序 故程序未对找到的路径进行保存，找到正确路径后直接输出 */ struct node{       int weight;       int child;       int bro;}; node tree[105];int length; //要求的路径长度 /*用后序遍历的形式访问Tree当访问到叶子节点时,输出数组 */ int s[150];//简易stack int site;//stack的指针(数组长度) void push(int* s,int value){     s[site++]=value;}void pop(int* s){     --site;}/*栈操作到此结束 */ void postOrder(int root){    if(root==-1)//找到叶子节点     {       int result=0;       for(int i=0;i<site;++i)                 result+=tree[s[i]].weight;       if(length==result)       {           for(int i=0;i<site-1;++i)              cout<<tree[s[i]].weight<<" ";          cout<<tree[s[site-1]].weight<<endl;       }       return;    }    else if(root==-2)//区分bro为空的情况        return ;    push(s,root);    postOrder(tree[root].child);    pop(s);    postOrder(tree[root].bro);} int cmp(int a,int b) {     return tree[a].weight>tree[b].weight; }int main(){    for(int i=0;i<105;++i)//Tree初始化     {        tree[i].child=-1;        tree[i].bro=-2;    }    int nodeNum,noneLeafNum;//节点数，非叶子节点数     cin>>nodeNum>>noneLeafNum>>length;    for(int i=0;i<nodeNum;++i)//节点的权值         cin>>tree[i].weight;    int father;    int numOfChild;    int children[105];//暂存某个节点输入的孩子    for(int i=0;i<noneLeafNum;++i)//构建Tree     {        cin>>father;        cin>>numOfChild;        if(numOfChild)        {            for(int j=0;j<numOfChild;++j)                cin>>children[j];            sort(children,children+numOfChild,cmp);//对父亲节点自己的孩子进行了递减排序             tree[father].child=children[0];            for(int j=1;j<numOfChild;++j)                tree[children[j-1]].bro=children[j];        }      }//建树完成     postOrder(0);//程序默认0作为root     //system("pause");    return 0;}