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Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47668 Accepted Submission(s): 25401
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> typedef long long ll; using namespace std; const int N=10000; int rank[N]; int parent[N]; int findset(int x) { if(x!=parent[x]) x=findset(parent[x]); return parent[x]; } void unionn(int root1,int root2) { int x=findset(root1); int y=findset(root2); if(x==y) return ; if(x<y) parent[x]=y; else if(y<=x) parent[y]=x; } int main() { int n,m; while(scanf("%d",&n)!=EOF) { if(n==0) { break; } int i; scanf("%d",&m); for(i=1;i<=n;i++) parent[i]=i; int x,y; for(i=1;i<=m;i++) { scanf("%d%d",&x,&y); unionn(x,y); } int ans=0; for(i=1;i<=n;i++) { if(parent[i]==i) ans++; } printf("%d\n",--ans); } }
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原文地址:http://www.cnblogs.com/Aa1039510121/p/5873871.html