1053. Path of Equal Weight (30)

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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

     今天被一道PAT顶级的题目折腾哭了，还好总算是做出来了，决定睡前还是发一篇博客吧，好久没写题解了，顶级的那道我明天再扯。

     这道题目其实水的很，要是我考甲级的时候30分的题目是这个，我做梦都会笑出来的。要说算法，其实根本没有算法，从顶点开始暴搜就好了，无需剪枝一路搜到叶子，如果到叶子的时候发现权重与题目要求的相符，就输出路径，为了维护字典序对每个节点的孩子进行一次排序，这样搜索的路径自动遵循字典序从小到大的原则。，然后，然后就没有什么了。

# include <iostream># include <vector># include <algorithm>using namespace std;struct Node {  int weight;  vector<int> edge;  int child;  Node():child(0){}  bool operator < (const Node& cmper) const  {    return weight > cmper.weight;  }};Node v[101];bool cmper(int a,int b){   return v[a] < v[b];}void PrintPath(vector<int>& path){   vector<int>::iterator it;   for (it = path.begin();it!=path.end();it++)      printf("%d%c",*it,it+1==path.end()?'\n':' ');}void dfs(int loca,int rest,vector<int>& path){//cout << "loca = " << loca << endl;   Node& cur = v[loca];   path.push_back(cur.weight);   rest -= cur.weight;   if (rest==0&&cur.child==0)      PrintPath(path);   else    {   vector<int>:: iterator it;   for (it = cur.edge.begin();it!=cur.edge.end();it++)       dfs(*it,rest,path);   }   path.pop_back();}int main(){  int i,j,k,root,temp;  int n,m,s;  cin >> n >> m >> s;  for (i=0;i<n;i++)    cin >> v[i].weight;  for (i=0;i<m;i++)    {  cin >> root >> k;  while (k--)    {  cin >> temp;          v[root].edge.push_back(temp);    }      v[root].child = k;}  for (i=0;i<n;i++)    sort(v[i].edge.begin(),v[i].edge.end(),cmper);  vector<int> path;  dfs(0,s,path);  return 0;}