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

N个结点，M个非叶节点，和为weight

N个结点，每个结点的值一行；

接下来M行非叶节点编号

                这个编号有几个子结点，接着几个子结点的编号

要求满足从root头到尾的总值为weight；当有多条时，按照同一水平的，值大的在前面

评测结果

时间结果得分题目语言用时(ms)内存(kB)用户8月12日 18:52答案正确301053C++ (g++ 4.7.2)2436datrilla

测试点

测试点结果用时(ms)内存(kB)得分/满分0答案正确143618/181答案正确14363/32答案正确14362/23答案正确14362/24答案正确13082/25答案正确23083/3

#include<iostream> #include<vector> #include<algorithm> using namespace std; class ROOTCMP{private:  vector<int>*weight;public:  ROOTCMP(vector<int>*wi):weight(wi){}  bool operator()(const int &A, const int&B)  { return (*weight)[A] >(*weight)[B]; }};int DuQUReadln(vector<int>*wi, vector<vector<int>>*nodes, int N, int M, vector<bool>*Noroot){  int index,temp;  for (index = 0; index < N; index++)    cin >> (*wi)[index];  while (M--)  {    cin >> index >> N;    while (N--)    {      cin >> temp;      (*Noroot)[temp] = true;      (*nodes)[index].push_back(temp);    }      sort((*nodes)[index].begin(), (*nodes)[index].end(), ROOTCMP(wi));  }  for (index = 0; index < N&&(*Noroot)[index]; index++);  return index;}void PATHrootDFS(vector<int>*wi, vector<vector<int>>*nodes, vector<int>*path, int Root, int countweight, int weight){  int newroot;  (*path).push_back((*wi)[Root]);  vector<int>::iterator iter = (*nodes)[Root].begin();  while (iter != (*nodes)[Root].end())  {    newroot = (*iter);    if ((*nodes)[newroot].size() != 0)    {      PATHrootDFS(wi, nodes, path, newroot, countweight + (*wi)[newroot], weight);      (*path).pop_back();    }    else if (countweight + (*wi)[newroot] == weight)    {      for (vector<int>::iterator b = (*path).begin(); b < (*path).end(); b++)      {        cout << (*b) << " ";      }      cout <<(*wi)[ newroot] << endl;    }    iter++;  }}int main(){      int N, M,weight,Root;     cin >> N >> M >> weight;  vector<int>wi(N);  vector<vector<int>>nodes(N);  vector<bool>Noroot(N, false);  vector<int>path;  Root=DuQUReadln(&wi, &nodes, N, M, &Noroot);     if (wi[Root] == weight)cout << weight << endl;  else  PATHrootDFS(&wi, &nodes, &path, Root, wi[Root], weight);  system("pause");  return 0;}