PAT A1053 Path of Equal Weight (30分)

题目链接：

题目描述
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.

输入
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.

输出
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​​ .

样例输入
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

样例输出
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int target;
struct NODE {
    int w;
    vector<int> child;
};
vector<NODE> v;
vector<int> path;
void dfs(int index, int nodeNum, int sum) {
    if(sum > target) return ;
    if(sum == target) {
        if(v[index].child.size() != 0) return;
        for(int i = 0; i < nodeNum; i++)
            printf("%d%c", v[path[i]].w, i != nodeNum - 1 ? ' ' : '\n');
        return ;
    }
    for(int i = 0; i < v[index].child.size(); i++) {
        int node = v[index].child[i];
        path[nodeNum] = node;
        dfs(node, nodeNum + 1, sum + v[node].w);
    }
    
}
int cmp1(int a, int b) {
    return v[a].w > v[b].w;
}
int main() {
    int n, m, node, k;
    scanf("%d %d %d", &n, &m, &target);
    v.resize(n), path.resize(n);
    for(int i = 0; i < n; i++)
        scanf("%d", &v[i].w);
    for(int i = 0; i < m; i++) {
        scanf("%d %d", &node, &k);
        v[node].child.resize(k);
        for(int j = 0; j < k; j++)
            scanf("%d", &v[node].child[j]);
        sort(v[node].child.begin(), v[node].child.end(), cmp1);
    }
    dfs(0, 1, v[0].w);
    return 0;
}