1053 Path of Equal Weight (30分)

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.

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<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.

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 {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​​.

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<iostream>
#include<vector>
#include<algorithm>

using namespace std;
struct node{
    int index;
    int weight;

};
bool cmp(node a, node b){

    return a.weight > b.weight;

}
vector<vector<node> > tree;
vector<int> vec;
vector<int> temp_path;
vector<vector<int> > path;
int n, m, s;
bool visit[110] = {false};
void dfs(int start, int sum){

    if(sum == s && tree[start].size() == 0){

        path.push_back(temp_path);
    }

    for(int i = 0; i < tree[start].size(); i++){
        if(!visit[tree[start][i].index]){
            visit[tree[start][i].index] = true;
            temp_path.push_back(tree[start][i].weight);
            dfs(tree[start][i].index, sum + tree[start][i].weight);
            temp_path.pop_back();
            visit[tree[start][i].index] = false;
        }
    }

}
int main(){
    
    
    cin >> n >> m >> s;
    vec.resize(n);
    tree.resize(n);
    for(int i = 0; i < n; i++){
        scanf("%d", &vec[i]);
    }
    for(int i = 0; i < m; i++){
        int id, k;
        scanf("%d %d", &id, &k);
        for(int i = 0; i < k; i++){
            int temp;
            scanf("%d", &temp);
            node temp_node;
            temp_node.weight = vec[temp];
            temp_node.index = temp;
            tree[id].push_back(temp_node);
        }
        sort(tree[id].begin(), tree[id].end(), cmp);
        

    }
    temp_path.push_back(vec[0]);
    visit[0] = true;
    dfs(0, vec[0]);

    for(int i = 0; i < path.size(); i++){
        printf("%d", path[i][0]);
        for(int j = 1; j < path[i].size(); j++){
            printf(" %d", path[i][j]);
        }
        printf("\n");
    }
    return 0;
}