1053. Path of Equal Weight (30)

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 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 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

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.

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

10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

深度遍歷dfs，dfs函數分兩部分，第一部分寫達到某些條件進行返回，第二部分寫遍歷下一個節點。

這道題在順序輸出上有兩個思路，一個是在children中添加子節點時，進行排序，使得在遍歷過程中實現順序遍歷；另一個是將遍歷後符合條件的結果放入一個數組中，最後在對數組排序，但是第二種方法，我在PAT上運行程序，有一個case顯示段錯誤，在牛客網上卻是正確的，我很詫異，可能是vector數組sort的問題？我不確定。

#include <bits/stdc++.h>

using namespace std;

int N, M, K;

struct Node {
    int weight;
    vector<int> children;
}nodes[102];

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

bool cmp2(vector<int> vec1, vector<int>vec2) {
    int len1 = vec1.size();
    int len2 = vec2.size();
    int len = min(len1, len2);
    for(int i = 0; i < len; i++) {
        if(nodes[vec1[i]].weight < nodes[vec2[i]].weight) {
            return false;
        }
        else if(nodes[vec1[i]].weight > nodes[vec2[i]].weight) {
            return true;
        }
    }
    return true;
}

vector<int> nodesVec;
//vector<int> nodesVecs[102];
//int INDEX = 0;

void dfs(int index, int weights) {
    //cout<< "index:"<< index<< endl;
    if(weights > K) {
        return;
    }
    if(nodes[index].children.size() == 0) {
        if(weights == K) {
            //nodesVecs[INDEX] = nodesVec;
            //INDEX++;
            for(int i = 0; i < nodesVec.size(); i++) {
                if(i != 0) cout<< " ";
                cout<< nodes[nodesVec[i]].weight;
            }
            cout<< endl;
        }
        return;
    }
    for(int i = 0; i < nodes[index].children.size(); i++) {
        nodesVec.push_back(nodes[index].children[i]);
        dfs(nodes[index].children[i], weights+nodes[nodes[index].children[i]].weight);
        nodesVec.erase(nodesVec.end()-1);
    }
}

int main()
{
    cin>>N>>M>>K;
    for(int i = 0; i < N; i++) {
        cin>> nodes[i].weight;
    }
    int index, s, c;
    for(int i = 0; i < M; i++) {
        cin>> index>> s;
        for(int j = 0; j < s; j++) {
            cin>> c;
            nodes[index].children.push_back(c);
        }
        sort(nodes[index].children.begin(), nodes[index].children.end(), cmp);
    }
    nodesVec.push_back(0);
    dfs(0, nodes[0].weight);
    /*
    sort(nodesVecs, nodesVecs+INDEX, cmp2);
    for(int i = 0; i < INDEX; i++) {
        for(int j = 0; j < nodesVecs[i].size(); j++) {
            if(j != 0) cout<< " ";
            cout<< nodes[nodesVecs[i][j]].weight;
        }
        cout<< endl;
    }*/
    return 0;
}

PAT 1053. Path of Equal Weight (30)