1053 Path of Equal Weight (30分)

1 题目

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<2^30, 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,⋯,An } is said to be greater than sequence {B1​​ ,B​2​​ ,⋯,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:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

2 解析

2.1 题意

给定一棵树和结点权值，求这颗树所有从根结点到叶子结点的路径权值之和等于S的路径，并按路径非递增的顺序输出。

2.2 思路

• 1，用树的静态写法创建二叉树
  • 结构体中：vector存放该结点所有孩子的编号,int类型变量存放权值
  • 为了 满足最后的权值从大到小排序输出， 在输入时，就把vector按vector的权值从大到小排序
• 2， 求路径就相当于对树进行先根遍历（DFS遍历），
  • 参数：
    • index：当前访问结点编号
    • numNode：当前路径path上的结点个数
    • weightSum:当前路径的权值和
  • 递归过程：
    • 1, weightSum > S时，直接return
    • 2, weightSum =S 时，若结点index为叶子结点时，return；否则，输出path数组中的所有数据
    • 3，若Sum< S时，说明还未满足条件。枚举当前index的所有子结点，对每个子结点child，将其存入path[numNode]中，然后在此基础上，进行下一层递归（child, numNode + 1, sum + Node[child.weight]）。

• 3 注意事项

• weightSum等于S时，必须判断该结点是叶子结点才输出，因为题目要求的是根结点到叶子结点的路径。
• cmp的所有情况都必须要有返回值，程序需要处理两条路径上结点权值完全相同的情况。

3 参考代码

• 方法一：

#include <cstdio>
#include <algorithm>

using std::sort;

int N, M, S;
const int MAXN = 110;
int path[MAXN];

struct node
{
    vector<int> child;
    int weight;
}Node[MAXN];

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

//index：当前访问的结点，numNode:当前路径path上的结点个数，weightSum：当前结点结点权值和
void DFS(int index, int numNode, int weightSum){
    if(weightSum > S) return;

    if(weightSum == S){
        //还没有到叶子结点返回
        if(Node[index].child.size() != 0) return;

        //到达叶子结点,path[]中存放了一个完整的路径，输出它
        for (int i = 0; i < numNode; ++i)
        {
            printf("%d", Node[path[i]].weight);

            if(i < numNode - 1){
                printf(" ");
            }else{
                printf("\n");
            }
        }

        return;
    }

    for (int i = 0; i < Node[index].child.size(); ++i)
    {
        int child = Node[index].child[i]; //结点index的第i个子结点编号
        path[numNode] = child;  //将结点child加入path末尾

        DFS(child, numNode + 1, weightSum + Node[child].weight); //递归进入下一层
    }
}

int main(int argc, char const *argv[])
{
    scanf("%d%d%d", &N, &M, &S);

    for (int i = 0; i < N; ++i)
    {
        scanf("%d", &Node[i].weight);
    }

    int ad, k;
    for (int i = 0; i < M; ++i)
    {
        scanf("%d%d", &ad, &k);

        int temp;
        for (int j = 0; j < k; ++j)
        {
            scanf("%d", &temp);
            Node[ad].child.push_back(temp);
        }

        sort(Node[ad].child.begin(), Node[ad].child.end(), cmp);
    }

    path[0] = 0;//路径第一个结点为零号结点

    DFS(0, 1, Node[0].weight);


    return 0;
}

• 方法二：
  用vector代替path数组

#include <cstdio>
#include <vector>
#include <algorithm>

using std::vector;
using std::sort;

int N, M, S;
const int MAXN = 110;

vector<int> path;

struct node
{
    vector<int> child;
    int weight;
}Node[MAXN];

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

//index：当前访问的结点，numNode:当前路径path上的结点个数，weightSum：当前结点结点权值和
void DFS(int index,  int weightSum){
    if(weightSum > S) return;

    if(weightSum == S){
        //还没有到叶子结点返回
        if(Node[index].child.size() != 0) return;

        //到达叶子结点,path[]中存放了一个完整的路径，输出它
        for (int i = 0; i < path.size(); ++i)
        {
            printf("%d", Node[path[i]].weight);

            if(i < path.size() - 1){
                printf(" ");
            }else{
                printf("\n");
            }
        }

        return;
    }

    for (int i = 0; i < Node[index].child.size(); ++i)
    {
        int child = Node[index].child[i]; //结点index的第i个子结点编号

        path.push_back(child);
        DFS(child, weightSum + Node[child].weight); //递归进入下一层
        path.pop_back();
    }
}

int main(int argc, char const *argv[])
{
    scanf("%d%d%d", &N, &M, &S);

    for (int i = 0; i < N; ++i)
    {
        scanf("%d", &Node[i].weight);
    }

    int ad, k;
    for (int i = 0; i < M; ++i)
    {
        scanf("%d%d", &ad, &k);

        int temp;
        for (int j = 0; j < k; ++j)
        {
            scanf("%d", &temp);
            Node[ad].child.push_back(temp);
        }

        sort(Node[ad].child.begin(), Node[ad].child.end(), cmp);
    }

    path.push_back(0);//路径第一个结点为零号结点

    DFS(0,  Node[0].weight);


    return 0;
}