PAT (Advanced Level) Practice 1053 Path of Equal Weight (30 分)

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<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_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:

我感觉这题目测试点还可以更难，这个好像只有一级相同，如果二级也相同呢。

#include <string>
#include <iostream>
#include <vector>
#include <algorithm>
#include <math.h>
#define max_size 101
using namespace std;
int n, m, s, total;
struct node
{int w;vector<int> sons;
};
node tree[max_size];
vector<int> path;
bool isRoot[max_size];
bool cmp(int a, int b)
{if (tree[a].w != tree[b].w){return tree[a].w > tree[b].w;}else{return false;}
}
void dfs(int root)
{total += tree[root].w;path.push_back(tree[root].w);if (total > s){total -= tree[root].w;path.pop_back();return;}else if (total == s){if (tree[root].sons.size() == 0)for (int i = 0; i < path.size(); i++){cout << path[i];if (i == path.size() - 1){cout << endl;}elsecout << " ";}total -= tree[root].w;path.pop_back();return;}sort(tree[root].sons.begin(), tree[root].sons.end(), cmp);for (int i = 0; i < tree[root].sons.size(); i++){dfs(tree[root].sons[i]);}total -= tree[root].w;path.pop_back();
}
int main()
{cin >> n >> m >> s;fill(isRoot, isRoot + n, true);for (int i = 0; i < n; i++){cin >> tree[i].w;}for (int i = 0; i < m; i++){int a, b, c;string sa, sc;cin >> sa >> b;a = (sa[0] - '0') * 10 + sa[1] - '0';for (int j = 0; j < b; j++){cin >> sc;c = (sc[0] - '0') * 10 + sc[1] - '0';tree[a].sons.push_back(c);isRoot[c] = false;}}int root;for (root = 0; root < n; root++){if (isRoot[root])break;}dfs(root);return 0;
}

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