A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.
Input Specification:
Each input file contains one test case. Each case starts with a line containing 0, the number of nodes in a tree, and M (<), the number of non-leaf nodes. 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 01.
The input ends with N being 0. That case must NOT be processed.
- Output Specification:
- For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.
The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1leaf node. Then we should output 0 1 in a line.
- Sample Input:
- 2 1 01 1 02
- Sample Output:
- 0 1
代码:
- #include <bits/stdc++.h>
- using namespace std;
- int N, M;
- vector<int> v[110];
- int vis[110];
- int ans[110];
- int deep = INT_MIN;
- void dfs(int st, int depth) {
- if(v[st].size() == 0) {
- ans[depth] ++;
- deep = max(deep, depth);
- }
- for(int i = 0; i < v[st].size(); i ++)
- dfs(v[st][i], depth + 1);
- }
- int main() {
- scanf("%d%d", &N, &M);
- memset(vis, 0, sizeof(vis));
- while(M --) {
- int p, k;
- scanf("%d%d", &p, &k);
- for(int i = 0; i < k; i ++) {
- int c;
- scanf("%d", &c);
- vis[c] = 1;
- v[p].push_back(c);
- }
- }
- dfs(1, 0);
- printf("%d", ans[0]);
- for(int i = 1; i <= deep; i ++)
- printf("%d", ans[i]);
- return 0;
- }
- dfs
- FH
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