【PAT 1053】 Path of Equal Weight 深度优先搜索

1053. Path of Equal Weight (30)

时间限制

10 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

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 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. 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, ..., 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.

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

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 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
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 fin

int 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 st;
	while(pa != -1){
		st.push(weight[pa]);
		pa = parent[pa];
	}
	cout< weight[bb];
}

int main()
{
	int n,m,s;
	scanf("%d %d %d",&n,&m,&s);
	int i,j;
	for(i=0;i