A1053. Path of Equal Weight
Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. 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 Wi (<1000) corresponds to the tree node Ti. 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, ..., An} is said to be greater than sequence {B1, B2, ..., Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for i=1, ... k, and Ak+1 > Bk+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
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<vector> 5 using namespace std; 6 typedef struct NODE{ 7 int weigh; 8 vector<int> child; 9 }node; 10 node tree[101]; 11 int N, M, S; 12 vector<int> ans; 13 bool cmp(int a, int b){ 14 return tree[a].weigh > tree[b].weigh; 15 } 16 void firstRoot(int root, int sum){ 17 sum += tree[root].weigh; 18 ans.push_back(root); 19 if(tree[root].child.size() == 0 && sum == S){ 20 int len = ans.size(); 21 for(int i = 0; i < len; i++){ 22 if(i == len - 1) 23 printf("%d\n", tree[ans[i]].weigh); 24 else printf("%d ", tree[ans[i]].weigh); 25 } 26 ans.pop_back(); 27 return; 28 } 29 if(sum > S || tree[root].child.size() == 0){ 30 ans.pop_back(); 31 return; 32 } 33 if(sum < S){ 34 int childnum = tree[root].child.size(); 35 for(int i = 0; i < childnum; i++) 36 firstRoot(tree[root].child[i], sum); 37 } 38 ans.pop_back(); 39 } 40 int main(){ 41 scanf("%d%d%d", &N, &M, &S); 42 for(int i = 0; i < N; i++) 43 scanf("%d", &tree[i].weigh); 44 int index, childn, tempc; 45 for(int i = 0; i < M; i++){ 46 scanf("%d%d", &index, &childn); 47 for(int j = 0; j < childn; j++){ 48 scanf("%d", &tempc); 49 tree[index].child.push_back(tempc); 50 } 51 sort(tree[index].child.begin(), tree[index].child.end(), cmp); 52 } 53 firstRoot(0, 0); 54 cin >> N; 55 return 0; 56 }
总结:
1、题意:给出一颗节点带权的树,给出指定权的和S,找出从根到叶节点的权的和为S的所有路径,如果有多条,则按照权的大小输出。
2、在写先根遍历的时候不要忘记push和pop,当在条件语句外加入当前节点时,需要在每个条件语句内部都pop,在整个函数结尾处也pop。 当找到符合条件的解时,不是将ans清空,而是仍然仅仅pop当前选择的元素。
3、要求从大到小输出,可以在读完每个节点时,对其孩子节点进行排序,权值大的孩子排在前面即可。