1053. Path of Equal Weight (30)

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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:

深搜当找到正确路径是记录，然后根据路径权排序，因为是树，所以不存在环，所以不需要记录是否访问过

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
#define MAX 105
using namespace std;
struct Node{
    int weight;
    int sum;
    vector<int> v1;
};

int n, m, target, counter = 0;
int weight[MAX] = {0};
int path[MAX] = {0};
vector<vector<int> > save(MAX);
Node tree[MAX] = {false};

void init()
{
    int father, son, num;
    for(int i=0; i<n; i++)
    {
        cin >> tree[i].weight;
    }
    for(int i=0; i<m; i++)
    {
        cin >> father >> num;
        for(int j=0; j<num; j++)
        {
            cin >> son;
            tree[father].v1.push_back(son);
        }
    }
}

void dfs(int n)
{
    if(weight[n]+tree[n].weight == target && tree[n].v1.empty())
    {
        int temp = n;
        while(temp != 0)
        {
            save[counter].push_back(tree[temp].weight);
            temp = path[temp];
        }
        counter++;
    }
    else if(weight[n]+tree[n].weight > target || tree[n].v1.empty())
    {
        return ;
    }
    vector<int>::iterator it;
    for(it=tree[n].v1.begin(); it!=tree[n].v1.end(); it++)
    {
        path[*it] = n;
        weight[*it] += weight[n] + tree[n].weight;
        dfs(*it);
    }
}

bool cmp(const vector<int> &a, const vector<int> &b)
{
    int i = a.size()-1;
    int j = b.size()-1;
    for(; i>=0 && j>=0; i--, j--)
    {
        if(a[i] != b[j])
            return a[i] > b[j];
    }
    if(i > j)
        return true;
    return false;
}
int main()
{
    cin >> n >> m >> target;
    init();
    dfs(0);
    sort(save.begin(), save.end(), cmp);
    for(int i=0; i<counter; i++)
    {
        cout << tree[0].weight;
        for(int j=save[i].size()-1; j>=0; j--)
        {
            cout << " " << save[i][j];
        }
        cout << endl;
    }
    return 0;

}