1053. Path of Equal Weight (30) -- 链表结构体的构造 和 dfs

1053. Path of Equal Weight (30)

Given a non-empty tree with root R, and with weight W_i assigned to each tree node T_i. Theweight 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 begreater than sequence {B₁, B₂, ..., 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

---

http://www.patest.cn/contests/pat-a-practise/1053

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <iterator>
using namespace std ;

#define INF  999999
#define N 105

int n ;
int m ;
int s ;

int weight[N] ;
vector<int> vt[N] ;

typedef struct node{
  int id ;
  int wi ;
  vector<struct node*> childs ;
  node(int _id , int _wi)
  {
    id = _id ;
    wi = _wi ;
    childs.clear() ;
  }
}Tree ;

bool cmp(Tree* t1 , Tree* t2)
{
  return t1->wi > t2->wi ;
}

Tree* createTree(int rid)
{
  Tree* root = new node(rid , weight[rid]);
  
  for(int i= 0;i<(int)vt[rid].size() ; i ++)
  {
    int child_id = vt[rid][i] ;
    root->childs.push_back(createTree(child_id));
  }
  return root ;
}

int fa[N] ;
int wn[N] ;

void dfs(Tree* rt)
{
  int len = rt->childs.size() ;
  if(wn[rt->id] == s && len == 0)
  {
    vector<int> vs ;
    vs.clear() ;
    int tmp = rt->id ;
    while(tmp != -1)
    {
      vs.push_back(weight[tmp]);
      tmp = fa[tmp];
    }
    int vslen = vs.size() ;
    for(int j = vslen - 1 ; j >= 1; j--)
      printf("%d ", vs[j]) ;
    printf("%d\n",vs[0]);
  }
  if(len > 1)
    sort(rt->childs.begin() , rt->childs.end() , cmp) ;
  for(int i= 0 ;i < len ; i++)
  {
    int child_id = rt->childs[i]->id ;
    fa[child_id] = rt->id ;
    wn[child_id] = weight[child_id] + wn[rt->id] ;
    dfs(rt->childs[i]);
  }
}  

int main()
{
  //freopen("in.txt","r",stdin);
  int i , j ;
  scanf("%d%d%d" , &n , &m , &s) ;
  for(i = 0 ; i< n ; i++)
  {
    scanf("%d",&weight[i]) ;
  }
  int id , k , tmpid ;
  for(i = 0 ; i < m ; i++)
  {
    scanf("%d%d",&id , &k);
    for(j = 0 ; j < k ; j ++)
    {
      scanf("%d",&tmpid);
      vt[id].push_back(tmpid);
    }
  }
  Tree* root = createTree(0) ;
  //pf(root);
  fa[0] = -1 ;
  wn[0] = weight[0] ;
  dfs(root);

  return 0 ;
}