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It’s an interesting experience to move from ICPC to work, end my college life and start a brand new journey in company.
As
is known to all, every stuff in a company has a title, everyone except
the boss has a direct leader, and all the relationship forms a tree. If
A’s title is higher than B(A is the direct or indirect leader of B), we
call it A manages B.
Now, give you the relation of a company, can you calculate how many people manage k people.
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <string> #include <vector> #include <queue> #include <math.h> #define eps 1e-8 #include <algorithm> using namespace std; int son[200]; int fa[200]; int main() { int n, k; int i, j; int u, v; while(scanf("%d %d", &n, &k)!=EOF){ memset(son, 0, sizeof(son));//出事哈每个人都有0个儿子 for(i=1; i<=n; i++){ fa[i]=i; //每个人的父亲是它自己 } for(i=1; i<n; i++){ scanf("%d %d", &u, &v); fa[v]=u; //将v的父亲确定为u son[u]++; //u的儿子数量+1 while( fa[u]!=u ){//如果u的父亲不是他自己 说明u不是根节点 u=fa[u]; //往上面回溯 直到根节点跳出 son[u]++; //经过的这些节点的儿子数全部+1 } } int cnt=0; for(i=1; i<=n; i++) if(son[i]==k) cnt++; printf("%d\n", cnt ); //遍历一下 输出 } return 0; }
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原文地址:http://www.cnblogs.com/yspworld/p/4684950.html