A1053. Path of Equal Weight

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_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:

 #include<cstdio>

 #include<iostream>

 #include<algorithm>

 #include<vector>

 using namespace std;

 typedef struct NODE{

     int weigh;

     vector<int> child;

 }node;

 node tree[];

 int N, M, S;

 vector<int> ans;

 bool cmp(int a, int b){

     return tree[a].weigh > tree[b].weigh;

 }

 void firstRoot(int root, int sum){

     sum += tree[root].weigh;

     ans.push_back(root);

     if(tree[root].child.size() ==  && sum == S){

         int len = ans.size();

         for(int i = ; i < len; i++){

             if(i == len - )

                 printf("%d\n", tree[ans[i]].weigh);

             else printf("%d ", tree[ans[i]].weigh);

         }

         ans.pop_back();

         return;

     }

     if(sum > S || tree[root].child.size() == ){

         ans.pop_back();

         return;

     }

     if(sum < S){

         int childnum = tree[root].child.size();

         for(int i = ; i < childnum; i++)

             firstRoot(tree[root].child[i], sum);

     }

     ans.pop_back();

 }

 int main(){

     scanf("%d%d%d", &N, &M, &S);

     for(int i = ; i < N; i++)

         scanf("%d", &tree[i].weigh);

     int index, childn, tempc;

     for(int i = ; i < M; i++){

         scanf("%d%d", &index, &childn);

         for(int j = ; j < childn; j++){

             scanf("%d", &tempc);

             tree[index].child.push_back(tempc);

         }

         sort(tree[index].child.begin(), tree[index].child.end(), cmp);

     }

     firstRoot(, );

     cin >> N;

     return ;

 }

总结：

1、题意：给出一颗节点带权的树，给出指定权的和S，找出从根到叶节点的权的和为S的所有路径，如果有多条，则按照权的大小输出。

2、在写先根遍历的时候不要忘记push和pop，当在条件语句外加入当前节点时，需要在每个条件语句内部都pop，在整个函数结尾处也pop。当找到符合条件的解时，不是将ans清空，而是仍然仅仅pop当前选择的元素。

3、要求从大到小输出，可以在读完每个节点时，对其孩子节点进行排序，权值大的孩子排在前面即可。

巴特西

A1053. Path of Equal Weight

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