Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3256 Accepted Submission(s): 1444
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.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 基础的并查集应用，主要注意的是路径压缩的迭代方式 #include<stdio.h> int parent[1005]; int rank[1005]; void Create(int x) { parent[x]=x; rank[x]=1; } int Find(int x) { int u=x; while(u!=parent[u]) { u=parent[u];//找根结点 } while(u!=x)//路径压缩 { int t=parent[x]; parent[x]=u; x=t; } return u; } void Union(int x,int y) { int l1=Find(x); int l2=Find(y); if(rank[l1]>rank[l2]) { parent[l2]=l1; } else { if(rank[l1]==rank[l2]) rank[l2]++; parent[l1]=l2; } } int main() { int n,m,t; int a,b; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) Create(i); for(int i=1;i<=m;i++) { scanf("%d%d",&a,&b); Union(a,b); } int count=0; for(int i=1;i<=n;i++) if(parent[i]==i) count++; printf("%d/n",count); } return 0; }
