PAT(Advanced)1053 Path of Equal Weight C++實現

PAT(Advanced)甲級1053 Path of Equal Weight C++實現

題目鏈接

1053 Path of Equal Weight

題目大意

給定一棵有N個節點和M個非葉子節點且所有節點都帶正整數權值的樹以及一個給定權值S，求解從根節點到葉子結點路徑累積權值爲S的所有路徑，並按照非遞增序列輸出

算法思路

根據樹的性質以及樹的層次遍歷定義，求解路徑累積權值爲給定值的序列，類似於圖的DFS，即樹的先序遍歷，從根節點遞歸到葉子節點，若未到達葉子節點但累積權值已經大於S，則往下的路徑無序繼續遍歷，而是回溯繼續查找其他路徑，若到達葉子節點符合要求，則存儲對應路徑序列，對於每個非葉子節點都需要遞歸地遍歷其所有孩子節點的對應路徑，得到結果後藉助sort()函數排序，用逆序迭代器逆序輸出即可得到非遞增序列

void preOrderTraverse(int p, vector<int> answer, int sum) {
    if (nonLeaves[p].size() == 0 && sum + weights[p] == S) {
    	// 葉子節點且符合要求
        answer.push_back(weights[p]);
        answers.push_back(answer);
        return;
    }
    for (int i = 0; i < nonLeaves[p].size(); i++) {
    	// 對於當前節點的所有孩子節點
        if (sum + weights[p] < S) {
        	// 若累積權值未大於S
            answer.push_back(weights[p]);
            preOrderTraverse(nonLeaves[p][i], answer, sum + weights[p]);
            // 回溯，查找其他路徑
            answer.pop_back();
        }
    }
}

AC代碼

/*
author : eclipse
email  : [email protected]
time   : Fri Jun 26 21:22:05 2020
*/
#include <bits/stdc++.h>
using namespace std;

int N, M, S;
vector<int> weights;
vector<vector<int> > nonLeaves;

vector<vector<int> > answers;


void preOrderTraverse(int p, vector<int> answer, int sum) {
    if (nonLeaves[p].size() == 0 && sum + weights[p] == S) {
        answer.push_back(weights[p]);
        answers.push_back(answer);
        return;
    }
    for (int i = 0; i < nonLeaves[p].size(); i++) {
        if (sum + weights[p] < S) {
            answer.push_back(weights[p]);
            preOrderTraverse(nonLeaves[p][i], answer, sum + weights[p]);
            answer.pop_back();
        }
    }
}

int main(int argc, char const *argv[]) {
    scanf("%d%d%d", &N, &M, &S);
    weights.resize(N);
     for (int i = 0; i < N; i++) {
        scanf("%d", &weights[i]);
    }
    nonLeaves.resize(N);
    for (int i = 0; i < M; i++) {
        int ID;
        int K;
        scanf("%d%d", &ID, &K);
        for (int j = 0; j < K; j++) {
            int child;
            scanf("%d", &child);
            nonLeaves[ID].push_back(child);
        }
    }
    vector<int> answer;
    preOrderTraverse(0, answer, 0);

    sort(answers.begin(), answers.end());

    for (vector<vector<int> >::reverse_iterator it = answers.rbegin() ; it != answers.rend(); it++) {
        vector<int>::iterator i = it->begin();
        printf("%d", *i++);
        for (; i != it->end(); i++) {
            printf(" %d", *i);
        }
        printf("\n");
    }
    return 0;
}

樣例輸入

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

樣例輸出

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

鳴謝

PAT

最後

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