PAT Advanced Level 1053 Path of Equal Weight
1053 Path of Equal Weight (30)(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 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 < 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:
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 {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:
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
/********************** author: yomi date: 18.8.16 ps: 本来简单dfs就可以了, 但是要求路径序列要按照非递增顺序输出 这个我想了好一会儿 甚至想到暂存到string数组里 然后sort 结果这是不现实的 除非再搞个哈希 麻烦 再看看题 其实就是每次往下搜索时先选权值最大的 所以 我把每一个父结点下的孩子们都按照 从大到小排了个序。成功解决。 **********************/ #include <iostream> #include <string> #include <vector> #include <algorithm> using namespace std; struct Node { int data; int id; vector<Node>child; }node[150]; bool vis[150]; long long int sum = 0, weight; vector<int>path; int cnt = 0; int cmp(Node a, Node b) { return a.data>b.data; } void dfs(int index) { if(node[index].child.size() == 0){ if(sum == weight){ for(int i=0; i<path.size()-1; i++){ cout << path[i] << ' '; } cout << path[path.size()-1] << endl; } return; } for(int i=0; i<node[index].child.size(); i++){ int id = node[index].child[i].id; if(!vis[id]){ sum += node[id].data; path.push_back(node[id].data); dfs(id); sum -= node[id].data; path.pop_back(); } } } int main() { int m, n, t, num, d; cin >> m >> n >> weight; for(int i=0; i<m; i++){ cin >> node[i].data; node[i].id = i; } for(int i=0; i<n; i++){ cin >> t >> num; for(int j=0; j<num; j++){ cin >> d; node[t].child.push_back(node[d]); } sort(node[t].child.begin(), node[t].child.end(), cmp); } sum += node[0].data; path.push_back(node[0].data); vis[0] = true; dfs(0); return 0; } /** 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 **/