PAT甲級1053 Path of Equal Weight (30 分)

1053 Path of Equal Weight (30 分)

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 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≤100, the number of nodes in a tree, M (<N), the number of non-leaf nodes, and 0<S<2​30​​, 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:

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 {A​1​​,A​2​​,⋯,A​n​​} is said to be greater than sequence {B​1​​,B​2​​,⋯,B​m​​} if there exists 1≤k<min{n,m} such that A​i​​=B​i​​ for i=1,⋯,k, and A​k+1​​>B​k+1​​.

Sample Input:

Sample Output:

      題目大意：

         給出有n個結點的樹以及m個非葉子結點以及目标值，下一行便是n個結點的權重，随後是m行，關于結點I 與 k個孩子編号。

         輸出權重從大到小序列。

思路：

         用dfs+領接表，并且錄入某結點的所有孩子結點時，讓權重大的結點排到前面。在dfs中遇到總值達到目标時就可輸出了。

參考代碼：

#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
vector<vector<int> > e;
vector<int> w, ans;
int n, m, s;
bool cmp(int a, int b){return w[a] > w[b];}
void dfs(int v, int weight){
	if(weight > s) return;
	ans.push_back(w[v]);
	if(weight == s && !e[v].size())
		for(int i = 0; i < ans.size(); ++i)	printf("%d%c", ans[i], i != ans.size() - 1? ' ':'\n');
	for(int i = 0; i < e[v].size(); ++i)
		dfs(e[v][i], weight + w[e[v][i]]);
	ans.pop_back();
}
int main(){
	scanf("%d%d%d", &n, &m, &s);
	w.resize(n), e.resize(n);
	for(int i = 0; i < n; ++i)	scanf("%d", &w[i]);
	for(int i = 0; i < m; ++i){
		int idx, k;
		scanf("%d%d", &idx, &k);
		e[idx].resize(k);
		for(int j = 0; j < k; ++j)	scanf("%d", &e[idx][j]);
		sort(e[idx].begin(), e[idx].end(), cmp);
	}
	dfs(0, w[0]);
	return 0;
}