1053 Path of Equal Weight (30 分)一般树的遍历 DFS+vector容器+sort排序

1053 Path of Equal Weight (30 分)

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<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=Bifor 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:

Special thanks to Zhang Yuan and Yang Han for their contribution to the judge's data.

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 110;
struct node {int weight;vector<int> child;
}Node[maxn];bool cmp(int a, int b) {return Node[a].weight > Node[b].weight;
}//对每个结点的数据域进行从大到小的排序，注意（）中是int不是node，return的是数组Node[].weight>Node[b].weightint n, m, S;//总结点数，非叶子结点数，给定的和
int path[maxn];//index为当前访问结点，numNode为当前路径上的结点个数
//sum为当前结点的权和void DFS(int index, int numNode, int sum) {if (sum > S)return;if (sum == S) {if (Node[index].child.size() != 0)return;//没到叶子结点//到达叶子结点，当前path[]中存放了一条完整的路径，输出他for (int i = 0; i < numNode; i++) {printf("%d", Node[path[i]].weight);if (i < numNode - 1)printf(" ");else printf("\n");}return;}for (int i = 0; i < Node[index].child.size(); i++) {int child = Node[index].child[i];path[numNode] = child;DFS(child, numNode + 1, sum + Node[child].weight);}
}int main() {scanf("%d%d%d", &n, &m, &S);for (int i = 0; i < n; i++) {scanf("%d", &Node[i].weight);}int id, k, child;for (int i = 0; i < m; i++) {scanf("%d%d", &id, &k);for (int j = 0; j < k; j++) {scanf("%d", &child);Node[id].child.push_back(child);}sort(Node[id].child.begin(), Node[id].child.end(), cmp);}path[0] = 0;DFS(0, 1, Node[0].weight);return 0;
}