本文主要是介绍【PAT 1053】 Path of Equal Weight 深度优先搜索，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
1053. Path of Equal Weight (30)
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 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 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 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 19Sample Output:
10 5 2 7 10 4 10 10 3 3 6 2 10 3 3 6 2
题意：
从给出的多叉树中，找到从根到叶的某一权值总和的路径，并根据节点权值从大到小输出。
分析：
使用Dijkstra算法，发现几条路径求出来以后，很难进行排序输出；
使用深度优先搜索，在搜索前根据权值进行从大到小排序，这样便保证了搜索的匹配结果顺序就是可直接输出的顺序。
代码：
#include <iostream>
#include <fstream>
#include <algorithm>
#include <stack>
#include <vector>
#include <queue>
#include <cstdio>
#include <cstring>
#include <memory>
using namespace std;//此代码使用前，需删除下面两行+后面的system("PAUSE")
ifstream fin("in.txt");
#define cin finint weight[101]={0};
int parent[101]={0};
int childNum[101]={0};
int sum[101]={0};
int linkNode[101][101]={0};void print(int pa){ //根据叶节点回溯打印链表stack<int> st;while(pa != -1){st.push(weight[pa]);pa = parent[pa];}cout<<st.top();st.pop();while(!st.empty()){cout<<' '<<st.top();st.pop(); }cout<<endl;
}void DFS(int cur,int s){ //深度优先搜索int next;for(int i=0;i<childNum[cur];i++){next = linkNode[cur][i];sum[next] = sum[cur]+weight[next];if(sum[next]==s){ //如果权值和等于目标值if(childNum[next]==0){ //又恰好是叶节点，则输出print(next);}else{ //不是叶节点，则跳出循环continue;}}else if(sum[next]<s){ //权值和小于目标，继续深度搜索DFS(next,s);}else{ //权值和大于目标，没必要继续深搜该节点continue;}}
}bool cmp(const int& aa,const int& bb){return weight[aa] > weight[bb];
}int main()
{int n,m,s;scanf("%d %d %d",&n,&m,&s);int i,j;for(i=0;i<n;i++)scanf("%d",&weight[i]);int nonleaf,num,leaf;parent[0] = -1;for(i=0;i<m;i++){scanf("%d %d",&nonleaf,&num);childNum[nonleaf] = num;for(j=0;j<num;j++){scanf("%d",&leaf);linkNode[nonleaf][j] = leaf;parent[leaf]=nonleaf;}sort(linkNode[nonleaf],linkNode[nonleaf]+num,cmp); //把每个节点的所有子节点从大到小排序}if(m==0){ //当仅有一个根节点时if(weight[0]==s)cout<<weight[0]<<endl;return 0;}memcpy(sum,weight,n*sizeof(int)); //拷贝权重到sum数组DFS(0,s);system( "PAUSE");return 0;
}这篇关于【PAT 1053】 Path of Equal Weight 深度优先搜索的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
