Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 44262 Accepted Submission(s): 23454
1 #include<iostream> 2 #include<string.h> 3 using namespace std; 4 5 int pre[1050]; 6 bool t[1050]; //t 用于标记独立块的根结点 7 8 int Find(int x) 9 { 10 int r=x; 11 while(r!=pre[r]) 12 r=pre[r]; 13 14 int i=x,j; 15 while(pre[i]!=r) 16 { 17 j=pre[i]; 18 pre[i]=r; 19 i=j; 20 } 21 return r; 22 } 23 24 void mix(int x,int y) 25 { 26 int fx=Find(x),fy=Find(y); 27 if(fx!=fy) 28 { 29 pre[fy]=fx; 30 } 31 } 32 33 int main() 34 { 35 int N,M,a,b,i,j,ans; 36 while(scanf("%d%d",&N,&M)&&N) 37 { 38 for(i=1;i<=N;i++) //初始化 39 pre[i]=i; 40 41 for(i=1;i<=M;i++) //吸收并整理数据 42 { 43 scanf("%d%d",&a,&b); 44 mix(a,b); 45 } 46 47 48 memset(t,0,sizeof(t)); 49 for(i=1;i<=N;i++) //标记根结点 50 { 51 t[Find(i)]=1; 52 } 53 for(ans=0,i=1;i<=N;i++) 54 if(t[i]) 55 ans++; 56 57 printf("%d\n",ans-1); 58 59 } 60 return 0; 61 }//dellaserss
原文:http://www.cnblogs.com/qinduanyinghua/p/5487545.html