PAT Advanced 1053 Path of Equal Weight (30) [树的遍历]

题目

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 diferent 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.

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 = Bi for 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 19

Sample Output:

10 5 2 7

10 4 10

10 3 3 6 2

10 3 3 6 2

题目分析

已知非叶子节点的所有子节点，已知每个节点的权重，已知一个权重和S，求权重和等于S的路径（该路径必须是从root到叶子节点）

注：多个满足条件的路径，必须非升序打印，序列A>序列B的条件为：A1_Ai与B1Bi相等，但是Ai+1>Bi+1

解题思路

1. 定义节点结构体（权重：w，子节点cds），int path[n]记录从root到当前节点路径
2. 题目要求权重非增序输出，每个节点的所有节点信息输入完成后，对所有子节点进行排序（权重由大到小，权重最大的节点排在最左边），深度优先遍历时会从最左边路径开始，可保证最后输出的路径满足非升序条件
3. dfs深度优先遍历树，参数numNode记录当前path中元素的个数，参数sum记录从root到当前节点的权重和
   
   3.1 若权重和>s，退出不再处理该路径
   
   3.2 若权重和==s
   
   3.2.1 若当前节点是叶子节点，打印路径
   
   3.2.2 若当前节点是非叶子节点，退出不再处理该路径

Code

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn=101;
int s,path[maxn];
struct node {
	int w;
	vector<int> cds;
} nds[maxn];
bool cmp(int a,int b) {
	return nds[a].w>nds[b].w;
}
// index当前处理节点在nds中的下标，numNode当前path数组中元素个数，sum从root到当前节点权重和
void dfs(int index, int numNode, int sum) {
	if(sum>s)return; //权重和超过s
	if(sum==s) {
		if(nds[index].cds.size()!=0)return; //权重和为s，但不是叶子节点
		//满足条件，权重和为s，且为叶子节点
		for(int i=0; i<numNode; i++) {
			if(i!=0)printf(" ");
			printf("%d",nds[path[i]].w);
			if(i==numNode-1)printf("\n");
		}
		return;
	}
	for(int i=0; i<nds[index].cds.size(); i++) {
		int cdi = nds[index].cds[i]; //子节点在nds中的下标
		path[numNode] = cdi;
		dfs(cdi, numNode+1, sum+nds[cdi].w);
	}
}
int main(int argc,char * argv[]) {
	int n,m,cn,id,cid;
	scanf("%d %d %d",&n,&m,&s);
	for(int i=0; i<n; i++) {
		scanf("%d",&nds[i].w);
	}
	for(int i=0; i<m; i++) {
		scanf("%d %d",&id,&cn);
		for(int j=0; j<cn; j++) {
			scanf("%d",&cid);
			nds[id].cds.push_back(cid);
		}
		sort(nds[id].cds.begin(),nds[id].cds.end(),cmp);
	}
	path[0]=0;
	dfs(0,1,nds[0].w);
	return 0;
}