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.
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.
For each test case, outputin a single line the number
of possible locations in the school the hamster may be found.
1 10 2 0 1 0 2 0 3 1 4 1 5 2 6 3 7 4 8 6 9
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1 //937MS 4188K 915 B G++
2 /*
3
4 有点暴力...水题。好像没用到什么算法
5
6 */
7 #include<iostream>
8 #include<vector>
9 #include<queue>
10 using namespace std;
11 struct node{
12 int id,s;
13 node(int x,int y){
14 id=x;s=y;
15 }
16 };
17 vector<node>V[100005];
18 int deal(int n,int d)
19 {
20 queue<node>Q;
21 int s=0;
22 Q.push(node(0,0));
23 while(!Q.empty()){
24 node t=Q.front();
25 Q.pop();
26 //printf("*%d %d\n",t.id,t.s);
27 if(t.s>d) s++;
28 for(int i=0;i<V[t.id].size();i++){
29 V[t.id][i].s=t.s+1;
30 Q.push(V[t.id][i]);
31 }
32 }
33 return s;
34 }
35 int main(void)
36 {
37 int t,n,d;
38 int x,y;
39 scanf("%d",&t);
40 while(t--)
41 {
42 scanf("%d%d",&n,&d);
43 for(int i=0;i<n;i++){
44 V[i].clear();
45 }
46 for(int i=1;i<n;i++){
47 scanf("%d%d",&x,&y);
48 V[x].push_back(node(y,0));
49 }
50 printf("%d\n",deal(n,d));
51 }
52 return 0;
53 }
