1053 Path of Equal Weight (30 分)

1053 Path of Equal Weight (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 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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where W​i​​ (<) 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 { is said to be greater than sequence { if there exists 1 such that A​i​​=B​i​​ for ,, 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这题挺不错的，但是算是一个dfs遍历的模板题，没加啥东西。 1 #include <bits/stdc++.h> 2 #define ll long long int 3 using namespace std; 4 int n,m,k; 5 struct Node 6 { 7 int to; 8 ll val; 9 friend bool operator < (const Node &a, const Node &b){ 10 return a.val > b.val; 11 } 12 }; 13 vector<Node> v[105]; 14 int vis[105]; 15 ll an[105]; 16 int dp[105]; 17 18 19 void output(int x){ 20 stack<int> st; 21 while(x != 0){ 22 st.push(an[x]); 23 x = dp[x]; 24 } 25 st.push(an[0]); 26 while(!st.empty()){ 27 cout << st.top(); 28 if(st.size() == 1){ 29 cout << endl; 30 }else{ 31 cout << " "; 32 } 33 st.pop(); 34 } 35 } 36 37 void dfs(int x, ll sum){ 38 vis[x] = 1; 39 if(sum == k){ 40 if(v[x].size() == 0) 41 output(x); 42 return ; 43 }else if(sum > k){ 44 return ; 45 } 46 for(int i = 0; i < v[x].size(); i++){ 47 if(vis[v[x][i].to]) 48 continue; 49 dp[v[x][i].to] = x; 50 dfs(v[x][i].to, sum+v[x][i].val); 51 } 52 } 53 54 55 int main(){ 56 scanf("%d%d%d", &n, &m, &k); 57 for(int i = 0; i < n; i++){ 58 cin >> an[i]; 59 } 60 int x,y,z; 61 for(int i = 0; i < m; i++){ 62 scanf("%d%d", &x,&y); 63 for(int j = 0; j < y; j++){ 64 scanf("%d", &z); 65 v[x].push_back({z,an[z]}); 66 } 67 } 68 for(int i = 0; i < n; i++){ 69 sort(v[i].begin(), v[i].end()); 70 } 71 dfs(0, an[0]); 72 return 0; 73 }

 

转载于:https://www.cnblogs.com/zllwxm123/p/11185839.html