HDOJ-----4707

Pet

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2507    Accepted Submission(s): 1233

Problem Description

One day, Lin Ji wake up in the morning and found that his pethamster escaped. He searched in the room but didn’t find the hamster. He tried to use some cheese to trap the hamster. He put the cheese trap in his room and waited for three days. Nothing but cockroaches was caught. He got the map of the school and foundthat there is no cyclic path and every location in the school can be reached from his room. The trap’s manual mention that the pet will always come back if it still in somewhere nearer than distance D. Your task is to help Lin Ji to find out how many possible locations the hamster may found given the map of the school. Assume that the hamster is still hiding in somewhere in the school and distance between each adjacent locations is always one distance unit.

 

Input

The input contains multiple test cases. Thefirst line is a positive integer T (0<T<=10), the number of test cases. For each test cases, the first line has two positive integer N (0<N<=100000) and D(0<D<N), separated by a single space. N is the number of locations in the school and D is the affective distance of the trap. The following N-1lines descripts the map, each has two integer x and y(0<=x,y<N), separated by a single space, meaning that x and y is adjacent in the map. Lin Ji’s room is always at location 0.

 

Output

For each test case, outputin a single line the number of possible locations in the school the hamster may be found.

 

Sample Input

110 20 10 20 31 41 52 63 74 86 9

 

Sample Output

2

一个人养的仓鼠跑了，他买了仓鼠食物等了三天，仓鼠也没回来，问仓鼠可能躲在哪，给出n个房间连通关系和能闻到食物的有效距离D，在有效距离内就会回来，但是等了三天没回来，说明不在距离内，食物始终在0号房间，就是给一棵树，每个点之间距离是单位距离，求出距离0号房间的距离大于D的点有几个，也不算是并查集，题目说没有回头路，可以从0房间走到任何房间，也就是一个有向无环图，所以输入两个数x，y，只能从x走到y（个人理解）

#include<cstdio>#include<cstring>#define maxn 110000int pre[maxn], vis[maxn], ans;int main(){    int t, m, n, x, y;scanf("%d", &t);    while(t--){    memset(pre, 0, sizeof(pre));    scanf("%d%d", &m, &n);ans = 0;for(int i = 0; i < m-1; i++){scanf("%d%d", &x, &y);pre[y] = pre[x]+1;if(pre[x] > n || pre[y] > n){ans++;}}printf("%d\n", ans);}    return 0;}