1053. Path of Equal Weight (30)

dfs函数携带vector形参记录搜索路径

时间限制

10 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 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 W_i (<1000) corresponds to the tree node T_i. 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, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor i=1, ... k, and 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

来源： <http://www.patest.cn/contests/pat-a-practise/1053>

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>

using namespace std;

map<int, vector<int>> mp;

vector<int> weight;
int n, m, aimWeight;
int pathWeight=0;

bool cmp(int a, int b) {
	return weight[a] > weight[b];
}

void dfs(int i,int w, vector<int> p) {
	p.push_back(i);

	if (mp[i].size() == 0) {
		if (w == aimWeight) {
			for (int j = 0; j < p.size(); j++) {
				cout << weight[p[j]];
				if (j != p.size() - 1)
					cout << " ";
				else
					cout << endl;
			}
		}
		p.clear();
		return;
	}

	else {
		for (int j = 0; j < mp[i].size(); j++)
			dfs(mp[i][j],weight[mp[i][j]]+w,p);
	}
}

int main(void) {	
	cin >> n >> m >> aimWeight;
	for (int i = 0; i < n; i++) {
		int wtemp;
		cin >> wtemp;
		weight.push_back(wtemp);
	}
	for (int i = 0; i < m; i++) {
		int id, k;
		cin >> id >> k;
		for (int j = 0; j < k; j++) {
			int idtemp;
			cin >> idtemp;
			mp[id].push_back(idtemp);
		}
		sort(mp[id].begin(), mp[id].end(), cmp);
	}
	vector<int> path;
	dfs(0,weight[0],path);

	return 0;
}

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