题目

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

分析

题目大意：输入一个有N个带权重节点，M个非叶子节点的树，给定一个S，求路径从根节点到叶子节点权重加起来等于S的所有路径，并按字典序降序输出。
思路：用结构体数组建立数据结构，并且在建立时对每个子节点按权重大小重新排序。dfs遍历的时候用数组path对路径进行跟踪，最后就直接输出path中0：childNode-1的权重。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
struct node{
    int w;
    int child_num;
    vector<int> child;
};
vector<int> path;
vector<node> tree;
int N, M, S;
void dfs(int r, int sum_w, int childNode)
{
    if(sum_w > S)
        return;
    if(sum_w == S)
    {
        if(tree[r].child_num != 0)
            return;
        for(int j = 0; j<childNode; j++)
        {
            cout << tree[path[j]].w;
            if(j != childNode-1)
                cout << " ";
            else
                cout << endl;
        }
    }
    for(int i = 0; i<tree[r].child_num; i++)
    {
        path[childNode] = tree[r].child[i];
        dfs(tree[r].child[i], sum_w+tree[tree[r].child[i]].w, childNode+1);
    }
}

int cmp(int a, int b)
{
    return tree[a].w > tree[b].w;
}

int main()
{

    cin >> N >> M >> S;
    tree.resize(N);
    path.resize(N);
    for(int i = 0; i<N; i++)
    {
        cin >> tree[i].w;
    }
    for(int i = 0; i<M; i++)
    {
        int ID;
        cin >> ID;
        cin >> tree[ID].child_num;
        tree[ID].child.resize(tree[ID].child_num);
        for(int j = 0; j<tree[ID].child_num; j++)
        {
            cin >> tree[ID].child[j];
        }
        sort(tree[ID].child.begin(), tree[ID].child.end(), cmp);
    }
//    for(int i = 0; i<N; i++)
//    {
//        cout << i << " " << tree[i].w << " " << tree[i].child_num << " ";
//        for(int j = 0; j<tree[i].child_num; j++)
//            cout << tree[i].child[j] << " ";
//        cout << endl;
//    }
    dfs(0, tree[0].w, 1);
    return 0;
}

总结

柳神的代码
学会如何在遍历树的时候对同一路径进行跟踪
利用sort，实现树按某一属性（权重）重新排序