HDU-4707-Pet【并查集】【搜索】

Pet

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

Total Submission(s): 2541    Accepted Submission(s): 1255

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

题意：给出 n 个房间，编号 0 到 n-1，又给出 n-1 个关系，问距 0 号距离大于 D 的房间有几个

代码1：

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int Q=1e5;
int N,D,room[Q+10];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(room,0,sizeof(room));
		scanf("%d %d",&N,&D);
		int a,b,ans=0;
		for(int i=1;i<N;i++)
		{
			scanf("%d %d",&a,&b);
			room[b]=room[a]+1;
			if(room[b]>D||room[a]>D)
			{
				ans++;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}

代码2：

#include<cstdio>
#include<cstring>
using namespace std;
const int Q=1e5;
int N,D,fa[Q+10];
int find(int x)  // 只向上查询祖先，查询根节点，不压缩 
{
	int y=x;
	while(y!=fa[y])
	{
		y=fa[y];
	}
	return y;
}
int jisuan(int x)  // 计算某个节点距离祖先的距离 
{
	int d=0;
	while(x!=fa[x])
	{
		x=fa[x];
		d++;
	}
	return d;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int a,b;
		scanf("%d %d",&N,&D);
		for(int i=0;i<N;i++)
		{
			fa[i]=i;
		}
		for(int i=1;i<N;i++)
		{
			scanf("%d %d",&a,&b);
			if(a>b)			// 这个 if else 是为了保证 0 为祖先，即保证 0 为根节点
				fa[a]=b;
			else
				fa[b]=a;
		}
		int ans=0;
		for(int i=0;i<N;i++)
		{
			if(find(i)==0)
			{
				if(jisuan(i)>D)
				{
					ans++;
				}
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}