原文地址:http://www.boiltask.com/blog/?p=1980

Pet

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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 1 10 2 0 1 0 2 0 3 1 4 1 5 2 6 3 7 4 8 6 9

Sample Output 2

这一题感觉并查集不是很好

代码似乎很容易出BUG

不过题目数据好像给的不是很严(还是很多的)

直接暴力深搜也会超时

如果dfs的话也需要先建树

用带权并查集的话主要是要注意

不能把0的父节点设为其它点

```
#include<stdio.h>
int par[100200],ran[100200];
int find(int m) {
if(m==par[m])
return m;
else {
ran[m]+=ran[par[m]];
return par[m]=find(par[m]);
}
}
void unite(int x,int y) {
if(x>y) {
int t=x;
x=y;
y=t;
}
int a=find(x);
int b=find(y);
if(x==y)
return ;
else {
par[b]=a;
ran[b]=ran[x]-ran[y]+1;
}
}
int main() {
int T;
scanf("%d",&T);
while(T--) {
int n,d;
scanf("%d %d",&n,&d);
for(int i=0; i<n; i++) {
ran[i]=0;
par[i]=i;
}
for(int i=1; i<n; i++) {
int a,b;
scanf("%d %d",&a,&b);
unite(a,b);
}
int res=0;
for(int i=0; i<n; i++) {
if(find(i)==0&&ran[i]>d)
res++;
}
printf("%d\n",res);
}
return 0;
}
```

参考博客:hdu 4707 Pet【带权并查集】 – mengxiang000000

题目地址:【杭电】[4707]Pet

原文地址:http://www.boiltask.com/blog/?p=1980