【HDU】-4707-Pet（并查集,好）

Pet

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2520    Accepted Submission(s): 1242

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

 

题意：求到 0 （根节点）距离大于 D 的个数

题解：不能路径压缩啊！不然所有的点到 0 的距离都变成 1 了！

不需要查找，直接合并；用 find() 判断根是不是为0，用 num()  判断距离大不于大于D。

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;#define CLR(a,b)  memset(a,b,sizeof(a))int f[100010];int sum;int n,k;int find(int x){while(x!=f[x])//注意这里不可以路径压缩 x=f[x];return x;}int num(int a){int ant=0;while(a!=f[a]){a=f[a];ant++;//找根，找一次，ant++ }return ant;}void fun(){for(int i=100010;i>0;i--){if(find(i)==0&&num(i)>k)//用find()判断根是不是0,用num()判断距离是不是大于 k sum++;}}int main(){int u;scanf("%d",&u);while(u--){sum=0;for(int i=0;i<100010;i++)f[i]=i;scanf("%d %d",&n,&k);int a,b;for(int i=0;i<n-1;i++){scanf("%d %d",&a,&b);f[b]=a;//不用查，直接合并 }fun();printf("%d\n",sum);} return 0;}