PAT 1053. Path of Equal Weight

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 = 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

分析

我觉得这道题还是很oK的，用邻接表方式储存子节点，并对每个父节点所有的子节点按权重排序，再用dfs，在dfs中保存路径，当dfs到叶节点时，即该节点的邻接表为空时，判断该条路径的和是否与题目要求的相等，相等则输出路径。注意在每次对一个节点的所有邻接表中的节点访问后，就要pop出，sum也要减去相应节点的weight，以维持路径和sum的正确性。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int weight[101],n,m,s,father,son,num,pre[101];
vector<int> list[101],res;
bool cmp(const int& n,const int& m){
	return weight[n]>weight[m];
}
void dfs(int root,int sum){
	res.push_back(weight[root]);
	sum+=weight[root];
	if(list[root].size()==0){
		if(sum==s){
			for(int i=0;i<res.size();i++)
			i>0?cout<<" "<<res[i]:cout<<res[i];
			cout<<endl;
		}
		sum-=res[res.size()-1];
		res.pop_back();
		return ;
	}
	for(int i=0;i<list[root].size();i++)
	    dfs(list[root][i],sum);
	sum-=res[res.size()-1];
	res.pop_back();
}
int main(){
	cin>>n>>m>>s;
	for(int i=0;i<n;i++)
	    cin>>weight[i];
	for(int i=0;i<m;i++){
		cin>>father>>num;
		for(int j=0;j<num;j++){
			cin>>son;
		    list[father].push_back(son);
		}
		sort(list[father].begin(),list[father].end(),cmp);
	}
	dfs(0,0);
	return 0;
}