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<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
思路：
步驟1:
一顆普通性質(zhì)樹，以結(jié)構(gòu)體node存放數(shù)據(jù)域和指針域，其中指針域使用vector存放孩子節(jié)點(diǎn)編號(hào)，考慮到最后輸出需要按權(quán)值從大到小排序，因此在數(shù)據(jù)讀入時(shí)就進(jìn)行節(jié)點(diǎn)排序。
步驟2：
用vector path對(duì)路徑進(jìn)行記錄，接下來進(jìn)行dfs，
如果weight>輸入的權(quán)重num，返回；
如果weight==輸入的權(quán)重num，進(jìn)行節(jié)點(diǎn)判斷，如果為葉子節(jié)點(diǎn)，輸出路徑，否則return；
如果weight<輸入的權(quán)重num，進(jìn)入下一層遞歸

#include<iostream> #include<vector> #include<algorithm> using namespace std; const int maxn = 101; int n, m, num; struct info {int data;//數(shù)據(jù)域，存儲(chǔ)節(jié)點(diǎn)的weightvector<int>v;//指針域，存儲(chǔ)孩子節(jié)點(diǎn) }node[maxn];//節(jié)點(diǎn)數(shù)組 vector<int>path; bool cmp(int a, int b)//節(jié)點(diǎn)數(shù)據(jù)域從大到小排序 {return node[a].data > node[b].data; } void dfs(int root,int weight) {if (weight > num)return;//當(dāng)前總重量大于numif (weight == num)//{if (node[root].v.size() > 0)return;//存在葉子節(jié)點(diǎn)for (int i = 0; i < path.size(); i++)//為葉子節(jié)點(diǎn){if (i == 0)cout << node[path[i]].data;else cout << " " << node[path[i]].data;}cout << endl;return;}for (int i = 0; i < node[root].v.size(); i++){path.push_back(node[root].v[i]);//用path存儲(chǔ)路徑dfs(node[root].v[i], weight + node[node[root].v[i]].data);path.pop_back();//回溯時(shí)記得彈出} } int main() {cin >> n >> m >> num;for (int i = 0; i < n; i++){cin >> node[i].data;}for (int i = 0; i < m; i++){int parent,k,child;cin >> parent >> k;for (int j = 0; j < k; j++){cin >> child;node[parent].v.push_back(child);}sort(node[parent].v.begin(), node[parent].v.end(),cmp);}path.push_back(0);//將節(jié)點(diǎn)0先存入路徑dfs(0, node[0].data); }

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