HDU:4707 Pet(并查集+某元素到根节点的距离计算)
2016-08-01 21:22
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Pet
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2522 Accepted Submission(s): 1244
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
For each test case, outputin a single line the number of possible locations in the school the hamster may be found.
[align=left]Sample Input[/align]
1
10 2
0 1
0 2
0 3
1 4
1 5
2 6
3 7
4 8
6 9
[align=left]Sample Output[/align]
2
[align=left]Source[/align]
2013 ACM/ICPC
Asia Regional Online —— Warmup
[align=left]Recommend[/align]
liuyiding
题目大意:给你n条边的信息,给个d,接下来n行为每条边的两端点。问距0这个节点距离大于d的元素个数有几个。
解题思路:让0作根节点,然后根据输入,将节点合并,然后遍历节点,往上搜根节点,每向上搜一层,那么步数(距离)肯定+1,直到搜到根节点0为止。
代码如下:
#include <cstdio> int pre[100010]; int sum[100010]; int ans; int find(int x) { int r=x; while(r!=pre[r]) { r=pre[r]; } return r; } int cnt(int x)//计算x到根节点的距离 { int sum=0; int r=x; while(r!=pre[r]) { r=pre[r]; sum++;//每向上搜索一次,步数+1 } return sum; } int main() { int t; scanf("%d",&t); for(int i=0;i<t;i++) { int n,d; for(int i=0;i<100010;i++)//初始化 { pre[i]=i; } scanf("%d%d",&n,&d); for(int i=0;i<n-1;i++) { int x,y; scanf("%d%d",&x,&y); if(x>y)//让小的作父节点 { int t=x; x=y; y=t; } pre[y]=x;//只并不查 } ans=0; for(int i=100010;i>=0;i--) { if(find(i)==0&&cnt(i)>d)//根节点是0,且距离0的距离大于d { ans++; } } printf("%d\n",ans); } return 0; }
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