A1053. Path of Equal Weight (30)

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_ifor 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:

10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <string>
#include <stack> 
#include <queue>
using namespace std;
const int maxn=110; 
int n,m,s;
struct node{
int weight;
vector<int> child; 
}Node[maxn]; 

int path[maxn];
bool cmp(int a,int b)
{
  return Node[a].weight>Node[b].weight;    
}
void DFS(int root,int num,int sum)
{
    if(sum>s)return;
    if(sum==s)
    {
        if(Node[root].child.size()==0)
        {
            //打印
            for(int i=0;i<num;i++)
            {
            printf("%d",Node[path[i]].weight);
            if(i<num-1)printf(" ");
            else printf("\n");    
            } 
        }else
        {
            return;
        }
    }
    for(int i=0;i<Node[root].child.size();i++)
    {
        int child=Node[root].child[i];
        path[num]=child;
        DFS(child,num+1,sum+Node[child].weight);
    }
}
int main(){
    scanf("%d %d %d",&n,&m,&s);
    for(int i=0;i<n;i++)
    {
        int tmp;
        scanf("%d",&tmp);
        Node[i].weight=tmp;
    }
    for(int i=0;i<m;i++)
    {
        int id,k,child;
        scanf("%d %d",&id,&k);
        for(int j=0;j<k;j++)
        {
            scanf("%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;
}