PAT甲级 -- 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=Bi for 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.

我的思路：

1. 建立结构体，存储权值，还有孩子。

2. 进行dfs，中间要加入权重值和的计算判断。

3. 进行代码实现的时候，dfs又把自己绕晕进去...我好想哭啊，递归怎么这么难写...下午按照柳神的思路再写一遍把！

柳神：

1. 在录入树的结点的时候就进行排序，以便深搜的时候直接从大到小输出。第一次见到这种cmp函数的书写方式诶，感觉学到了！！

2. 还有就是控制输出那个地方，也很新奇，值得学习！

3. path数组，传入一个nodeNum记录对当前路径来说这是第几个结点（这样直接在path[nodeNum]里面存储当前结点的孩子结点的序号，这样可以保证在先判断return的时候，path是从0~numNum-1的值确实是要求的路径结点）。然后每次要遍历下一个孩子结点的之前，令path[nodeNum] = 孩子结点的序号，这样就保证了在return的时候当前path里面从0~nodeNum-1的值就是要输出的路径的结点序号，输出这个序号的权值即可，直接在return语句里面输出。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int target;
struct NODE {int w;vector<int> child;
};
vector<NODE> v;
vector<int> path;
void dfs(int index, int nodeNum, int sum) {if(sum > target) return ;if(sum == target) {if(v[index].child.size() != 0) return;  //没有到叶子结点，返回for(int i = 0; i < nodeNum; i++) //到了叶子结点并且权值一样，进行输出printf("%d%c", v[path[i]].w, i != nodeNum - 1 ? ' ' : '\n');return ;}for(int i = 0; i < v[index].child.size(); i++) {  //递归访问孩子结点int node = v[index].child[i];path[nodeNum] = node;dfs(node, nodeNum + 1, sum + v[node].w);}}
int cmp1(int a, int b) {return v[a].w > v[b].w;
}
int main() {int n, m, node, k;scanf("%d %d %d", &n, &m, &target);v.resize(n), path.resize(n);for(int i = 0; i < n; i++)scanf("%d", &v[i].w);for(int i = 0; i < m; i++) {scanf("%d %d", &node, &k);v[node].child.resize(k);for(int j = 0; j < k; j++)scanf("%d", &v[node].child[j]);sort(v[node].child.begin(), v[node].child.end(), cmp1);}dfs(0, 1, v[0].w);return 0;
}