837. 连通块中点的数量 AcWing
作者:互联网
并查集模板题
当两个点互相可达,我们称它们连通.本题判断连通点的个数,就是判断同一集合下点的个数,一棵树下子节点各不同,因此用数组sizes记录下标为根节点的点的个数
1 #include <iostream> 2 using namespace std; 3 const int N = 1e5+10; 4 int p[N],sizes[N]; 5 int find(int x) 6 { 7 if(p[x]!=x) p[x] = find(p[x]); 8 return p[x]; 9 } 10 int main() 11 { 12 int n,m; 13 char op[3]; 14 scanf("%d%d",&n,&m); 15 for(int i=1;i<=n;i++) { p[i] = i;sizes[i] =1; } 16 while(m--){ 17 scanf("%s",op); 18 if(op[0]=='C') { 19 int x,y; 20 scanf("%d%d",&x,&y); 21 if(find(x)==find(y)) continue; 22 sizes[find(y)]+=sizes[find(x)]; 23 p[find(x)] = find(y); 24 } 25 else if(op[0]=='Q'){ 26 if(op[1]=='1'){ 27 int x,y; 28 scanf("%d%d",&x,&y); 29 if(find(p[x])==find(p[y])) printf("Yes\n"); 30 else printf("No\n"); 31 }else if(op[1]=='2'){ 32 int x; scanf("%d",&x); 33 printf("%d\n",sizes[find(x)]); 34 } 35 } 36 } 37 return 0; 38 }
标签:10,连通,837,sizes,int,个数,中点,find,AcWing 来源: https://www.cnblogs.com/newblg/p/14224747.html