PAT 1053. Path of Equal Weight (30)

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Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The weight of a path from R to Lis 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.

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 Wi (<1000) corresponds to the tree node Ti. 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, …, An} is said to be greater than sequence {B1, B2, …, Bm} if there exists 1 <= k < min{n, m} such that Ai = Bifor i=1, … k, and Ak+1 > Bk+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

树的遍历，用string保存它的路径，用深搜去求长度

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

using namespace std;

int *weight;
vector<vector<int> > tree;

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

void dfs(int node, string path, int s) {
    if (tree[node].size() == 0 && s == weight[node]) {
        path += " " + to_string(weight[node]);
        cout << path << endl;
        return;
    }
    for (int i = 0; i < tree[node].size(); i++) {
        dfs(tree[node][i], path + " " + to_string(weight[node]), s - weight[node]);
    }
}

int main() {
    int n = 0, m = 0, s = 0;
    scanf("%d %d %d", &n, &m, &s);
    weight = new int[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &weight[i]);
    }
    tree.resize(n);
    for (int i = 0; i < m; i++) {
        int id = 0, k = 0;
        scanf("%d %d", &id, &k);
        tree[id].resize(k);
        for (int j = 0; j < k; j++) {
            scanf("%d", &tree[id][j]);
        }
        sort(tree[id].begin(), tree[id].end(), cmp);
    }
    if (tree[0].size() == 0 && weight[0] == s) printf("%d", s);
    for (int i = 0; i < tree[0].size(); i++) {
        string path = to_string(weight[0]);
        dfs(tree[0][i], path, s - weight[0]);
    }
    delete[] weight;
    return 0;
}