hd4707 Pet

Pet

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1782    Accepted Submission(s): 860

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
 
嗯，题目大一就是说，求与根节点距离超过一个给定数值的子节点的个数。这里第一个数字是测试数据个数，然后n和d，n是节点个数从0到n-1，d是给定的距离。之后是相连的节点并且距离为1.这里根节点就一个是0.
开始一看与并查集有关，就直接写find函数和join函数，后来再一看，汗，根本不需要，只要一句连一下就ok。看代码吧。
#include<cstdio>#include<cstring>int map[100010];void join(int a,int b){    map[b]+=map[a];//就是这一句~}int main(){    int i,t,n,d,x,y,cnt;    scanf("%d",&t);    while(t--)    {        cnt=0;        scanf("%d%d",&n,&d);        for(i=0;i<n;++i)        {            map[i]=1;        }        map[0]=0;        for(i=1;i<n;++i)        {            scanf("%d%d",&x,&y);            join(x,y);            }        for(i=0;i<n;++i)        {            if(map[i]>d)                cnt++;        }        printf("%d\n",cnt);    }    return 0;}