1053 Path of Equal Weight-PAT甲级

Given a non-empty tree with root RRR, and with weight WiW_iW​i​​ assigned to each tree node TiT_iT​i​​. The weight of a path from RRR to LLL is defined to be the sum of the weights of all the nodes along the path from RRR to any leaf node LLL.

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<N≤1000 < N \le 1000<N≤100, the number of nodes in a tree, MMM (<N< N<N), the number of non-leaf nodes, and 0<S<2300 < S < 2^{30}0<S<2​30​​, the given weight number. The next line contains NNN positive numbers where WiW_iW​i​​ (<1000<1000<1000) corresponds to the tree node TiT_iT​i​​. Then MMM 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}\{A_1, A_2, \cdots , A_n\}{A​1​​,A​2​​,⋯,A​n​​} is said to be greater than sequence {B1,B2,⋯,Bm}\{B_1, B_2, \cdots , B_m\}{B​1​​,B​2​​,⋯,B​m​​} if there exists 1≤k<min{n,m}1 \le k < min\{n, m\}1≤k<min{n,m} such that Ai=BiA_i = B_iA​i​​=B​i​​ for i=1,⋯,ki=1, \cdots , ki=1,⋯,k, and Ak+1>Bk+1A_{k+1} > B_{k+1}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

考察树的遍历以及存储

考察树的路径上从根节点到某个叶子结点的权值和为某个固定的值，并按照字典序的顺序从大到小的输出

为了满足题目中从大到小的输出，我们可以在输入每个结点的字结点的权值时将其排序，从而访问的时候便是从大到小，利用dfs'对树的路径进行访问，遇到为叶子结点并且当前的权值和为给定的值，便输出此种解

满分代码如下；

#include<bits/stdc++.h>
using namespace std;
const int maxn=110;
int path[maxn];
struct node{
	int w;//权重 
	vector<int>child;//孩子结点 
}nd[maxn];
int n,m,s;
bool cmp(int a,int b){
	return nd[a].w>nd[b].w;//按照从大到小排序 
}
void dfs(int root,int sum,int layer){
	if(sum>s){//递归的边界
		return;
	}
	if(sum==s&&nd[root].child.size()!=0)
		return;
	if(sum==s){
		for(int i=0;i<layer;i++){
			if(i!=0) printf(" ");
			printf("%d",nd[path[i]].w);
		}
		printf("\n");
		return;
	}
	for(int i=0;i<nd[root].child.size();i++){
		int child=nd[root].child[i];
		path[layer]=child;
		dfs(child,sum+nd[child].w,layer+1);
	}
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>m>>s;
	for(int i=0;i<n;i++)
		cin>>nd[i].w;
	for(int i=1;i<=m;i++){
		int id,k,son;
		cin>>id>>k;
		for(int j=1;j<=k;j++){
			cin>>son;
			nd[id].child.push_back(son);
		}
		sort(nd[id].child.begin(),nd[id].child.end(),cmp);
	}
	path[0]=0;
	dfs(0,nd[0].w,1);
	return 0;
}