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[HTML_REMOVED]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<cstring>
#include<vector>
#include<algorithm>

using namespace std;

const int N=110;

int n,m,S;
int w[N];  //结点权重 
bool g[N][N];  //邻接矩阵存储图 
vector<vector<int>> ans;  //存答案 

void dfs(int u,int s,vector<int>&path)
{
    bool is_leaf=true;
    for(int i=0;i<n;i++)
        if(g[u][i])
        {
            is_leaf=false;
            break;
        }
    if(is_leaf)
    {
        if(s==S) ans.push_back(path);
    }
    else
    {
        for(int i=0;i<n;i++)
            if(g[u][i])
            {
                path.push_back(w[i]);
                dfs(i,s+w[i],path);
                path.pop_back();   //恢复现场 
            }
    }
}
int main()
{
    cin>>n>>m>>S;
    for(int i=0;i<n;i++) cin>>w[i];

    while(m--)
    {
        int id,k;
        cin>>id>>k;
        while(k--)
        {
            int son;
            cin>>son;
            g[id][son]=true;
        }
    }

    //vector<int> path({w[0]});
    vector<int> path;
    path.push_back(w[0]);  //存过程结果 

    dfs(0,w[0],path);

    sort(ans.begin(),ans.end(),greater<vector<int>>());  //按字典序排序 

    for(auto p:ans)
    {
        cout<<p[0];
        for(int i=1;i<p.size();i++) cout<<' '<<p[i];
        cout<<endl;
    }

    return 0;
}