How Many Tables//并查集
Posted w-j-c
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了How Many Tables//并查集相关的知识,希望对你有一定的参考价值。
题目:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41763 Accepted Submission(s): 20915
Problem Description
Today
is Ignatius‘ birthday. He invites a lot of friends. Now it‘s dinner
time. Ignatius wants to know how many tables he needs at least. You have
to notice that not all the friends know each other, and all the friends
do not want to stay with strangers.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.
For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
Input
The
input starts with an integer T(1<=T<=25) which indicate the
number of test cases. Then T test cases follow. Each test case starts
with two integers N and M(1<=N,M<=1000). N indicates the number of
friends, the friends are marked from 1 to N. Then M lines follow. Each
line consists of two integers A and B(A!=B), that means friend A and
friend B know each other. There will be a blank line between two cases.
Output
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
Sample Input
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
思路:
代码:
//按秩合并的并查集 #include <iostream> int Find(int a[], int x)//找根 { while(x != a[x]) { x = a[x]; } return x; } bool IsConnected(int a[], int x, int y)//判断是否已连接 { return Find(a,x) == Find(a,y); } void Union(int a[], int b[], int x, int y)//合并两个树 { int c = Find(a,x); int d = Find(a,y); if(b[c] < b[d])//小的树连到大的树上 a[c] = d; else { a[d] = c; if(b[c] == b[d])//若两个树高度相等,则合并后高度+1 b[c]++; } } int main() { using namespace std; int T; cin >> T; while(T--) { int N, M; int x, y; cin >> N >> M; int num = N; int * a = new int[N+1]; int * b = new int[N+1];//记录树的高度 for(int i = 0; i < N+1; i++) { a[i] = i; b[i] = 1; } while(M--) { cin >> x >> y; if(IsConnected(a,x,y)) continue; Union(a,b,x,y); num--; } printf("%d ", num); } }
以上是关于How Many Tables//并查集的主要内容,如果未能解决你的问题,请参考以下文章
HDU - 1213 How Many Tables(并查集)