题意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，使得每条路径上的结点的权值之和等于给定的常数S。如果有多条这样的路径，则按路径非递增的顺序输出。其中路径的大小是指，如果两条路径分别为

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int MAXN =110;
struct node{
    int weight;
    vector<int>child;
}Node[MAXN];
bool cmp(int a,int b){
    return Node[a].weight>Node[b].weight;
}
int n,m,S;
int path[MAXN];
void DFS(int index,int numNode,int sum){
    if(sum>S)return;
    if(sum==S){
        if(Node[index].child.size()!=0) return;
        for(int i=0;i<numNode;i++){
            printf("%d",Node[path[i]].weight);
            if(i<numNode-1) printf(" ");
            else printf("\n");
        }
        return ;
    }
    for(int i=0;i<Node[index].child.size();i++){
        int child=Node[index].child[i];
        path[numNode]=child;
        DFS(child,numNode+1,sum+Node[child].weight);
    }
}
int main(){
    scanf_s("%d%d%d",&n,&m,&S);
    for(int i=0;i<n;i++){
        scanf_s("%d",&Node[i].weight);
    }
    int id,k,child;
    for(int i=0;i<n;i++){
        scanf_s("%d%d",&id,&k);
        for(int j=0;j<k;j++){
            scanf_s("%d",&child);
            Node[id].child.push_back(child);
        }
        sort(Node[id].child.begin(),Node[id].child.end(),cmp);
    }
    path[0]=0;
    DFS(0,1,Node[0].weight);
    return 0;
}