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

Special thanks to Zhang Yuan and Yang Han for their contribution to the judge's data.

 

C++:

/*
 @Date    : 2018-08-28 20:41:41
 @Author  : 酸饺子 ([email protected])
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://www.patest.cn/contests/pat-a-practise/1053
 */

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

using namespace std;

const int MAXN = 101;
int N, M, S;

struct TreeNode
{
    int weight;
    vector<int> children;
};

TreeNode T[MAXN];

vector<int> thisPath;
vector<vector<int>> resPaths;
int thisW = 0;

void dfs(int r)
{
    thisPath.push_back(T[r].weight);
    thisW += T[r].weight;
    if (thisW == S)
    {
        if (T[r].children.empty())
        {
            resPaths.push_back(thisPath);
        }
    }
    else if (thisW < S)
    {
        for (int c : T[r].children)
        {
            dfs(c);
        }
    }

    thisPath.pop_back();
    thisW -= T[r].weight;
    return;
}

int main()
{
    scanf("%d %d %d", &N, &M, &S);
    for (int i = 0; i != N; ++i)
        scanf("%d", &T[i].weight);
    for (int i = 0; i != M; ++i)
    {
        int id, k;
        scanf("%d %d", &id, &k);
        for (int j = 0; j != k; ++j)
        {
            int c;
            scanf("%d", &c);
            T[id].children.push_back(c);
        }
    }

    dfs(0);

    sort(resPaths.rbegin(), resPaths.rend());
    for (auto &path : resPaths)
    {
        int output = 0;
        for (int p : path)
        {
            if (output++)
                putchar(' ');
            printf("%d", p);
        }
        putchar('\n');
    }
    return 0;
}