【PAT（甲级）】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 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

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≤kBk+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

解题思路：

从树的根开始，走到底部，如果路径上所有节点的权重加起来刚好等于S，则输出这条路径。

题目意思很清晰，所以只要dfs()这颗树，把节点的权重加一下就可以解决这道题的大部分测试点了。主要的问题在于要求输出的顺序是先输出节点权重大的，再输出节点权重小的。我利用了vector 来存储路径，所以直接用了sort和自己写的cmp来对结果进行重新排序。

最后直接输出节点权重即可。

易错点：

1. 最后一个测试点是输出的顺序不符合题意。如果你对路径的排序是一层一层的排序的话（就是直接对节点进行sort），最后输出的结果就会有错，一定要把所有可行的路径都存储进一个结构中，在对他们一起排序才行。

例如下面这颗树：

如果只是对层来排序的话，输出就是

10 7 2 8

10 7 8 2

实际上我们需要的输出是

10 7 8 2

10 7 2 8 

代码：

#include
using namespace std;
int N;//树的节点数
int M;//非叶子节点的个数
int S;//总权重
int T[101];//节点对应的权重
vector<int> tree[101];//存储树的孩子，下标为根节点，里面为孩子 
 
vector<int> tpath;//存储路径 
vectorint>> path;//存储所有路径 
 
void dfs(int a,int sum){
	if(sum>S) return;
	if(sum == S){
		if(tree[a].size()!=0){
			return;
		}
		path.push_back(tpath);
		return;
	}
	if(tree[a].size()==0) return;//走到树底部时返回
	
	for(int i=0;isize();i++){
		tpath.push_back(T[tree[a][i]]);
		dfs(tree[a][i],sum+T[tree[a][i]]);
		tpath.pop_back();
	}
	
}
 
bool cmp(vector<int> a,vector<int> b){//比较权重大小 
    return a>b;
}
 
int main(){
	cin>>N>>M>>S;
 
	for(int i=0;i
		cin>>T[i];
	} 
	
	for(int i=0;i
		int t,link;
		cin>>t>>link;
		for(int j=0;j
			int num;
			cin>>num;
			tree[t].push_back(num);
		}
	}
	
	tpath.push_back(T[0]);//将根节点先放入路径中 
	dfs(0,T[0]);
 	sort(path.begin(),path.end(),cmp);
//    sort(path.begin(), path.end(), greater>());
	for(int i=0;isize();i++){
		cout<0];
		for(int j=1;jsize();j++){
			cout<<" "<
		}
		cout<
	}
	
	
	return 0;
}