PAT_A1053#Path of Equal Weight

Source:

PAT A1053 Path of Equal Weight (30 分)

Description:

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 Rto 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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where W​i​​ (<) 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 { is said to be greater than sequence { if there exists 1 such that A​i​​=B​i​​ for ,, 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

Keys:

• 二叉树的建立和遍历

Code:

/*
time: 2019-06-28 16:28:41
problem: PAT_A1053#Path of Equal Weight
AC: 21:32

题目大意：
树的带权路径长度：根结点到各个叶子结点带权路径长度之和
现给定带权路径长度，打印所有等于该长度的路径

输入：
第一行给出：结点数N<100，分支结点数M，给定带权路径长度S
接下来一行，N个结点的权值（0~n-1）
接下来M行，分支结点id，孩子数K，各孩子id（根结点00）
输出：
各路径按照权值非增打印

基本思路：
递减存储各分支结点的孩子结点，遍历并统计带权路径，符合条件的输出之
*/
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int M=110;
vector<int> tree[M],path;
int w[M],pt=0,s;

bool cmp(int a, int b)
{
    return w[a] > w[b];
}

void Travel(int root, int value)
{
    if(tree[root].size()==0 && value==s)
    {
        path.push_back(root);
        for(int i=0; i<path.size(); i++)
            printf("%d%c", w[path[i]],i==path.size()-1?'\n':' ');
        path.pop_back();
        return;
    }
    path.push_back(root);
    for(int i=0; i<tree[root].size(); i++)
        Travel(tree[root][i],value+w[tree[root][i]]);
    path.pop_back();
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE

    int n,m;
    scanf("%d%d%d", &n,&m,&s);
    for(int i=0; i<n; i++)
        scanf("%d", &w[i]);
    for(int i=0; i<m; i++)
    {
        int id,k,kid;
        scanf("%d%d", &id, &k);
        for(int j=0; j<k; j++)
        {
            scanf("%d", &kid);
            tree[id].push_back(kid);
        }
        sort(tree[id].begin(), tree[id].end(), cmp);
    }
    Travel(0,w[0]);

    return 0;
}