1053 Path of Equal Weight （30 分）（树的遍历）

给定一棵树和每个结点的权值，求所有从根结点到叶子结点的路径，让每条路径上的结点的权值之和等于给定的常数，如果有多条这样的路径，按照非递增的顺序输出

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n, m, s;
const int maxn = 110;
struct node {
    int weight;
    vector<int>child;
}Node[maxn];
vector<int>ans;
bool comp(int a, int b)
{
    return Node[a].weight > Node[b].weight;
}
void DFS(int index, int sum)
{
    if (sum == s)
    {
        if (Node[index].child.size() != 0)return;
        for (int i = 0; i < ans.size(); i++)
        {
            printf("%d", Node[ans[i]].weight);
            if (i != ans.size() - 1)printf(" ");
        }
        printf("
");
        return;
    }
    if (sum > s)return;
    for (int i = 0; i < Node[index].child.size(); i++)
    {
        int child = Node[index].child[i];
        ans.push_back(child);
        DFS(child, sum + Node[child].weight);
        ans.pop_back();
    }
}
int main()
{
    scanf("%d%d%d", &n, &m, &s);
    for (int i = 0; i < n; i++)scanf("%d", &Node[i].weight);
    while (m--)
    {
        int id, k;
        scanf("%d%d", &id, &k);
        while (k--)
        {
            int idk;
            scanf("%d", &idk);
            Node[id].child.push_back(idk);
        }
        sort(Node[id].child.begin(), Node[id].child.end(), comp);
    }
    ans.push_back(0);
    DFS(0, Node[0].weight);
    return 0;
}