Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3563 Accepted Submission(s): 1072
/*** 题意:给出一些数,并给对这些数进行操作; ‘M a b‘代表把a堆加到b堆上面, ‘C a ‘ 代表查询当前点a下面有多少个 做法:并查集。 ***/ #include<iostream> #include<string.h> #include<algorithm> #include<stdio.h> #include<cmath> using namespace std; #define maxn 300000 + 3000 int fa[maxn]; ///记录父节点 int sum[maxn]; ///记录当前堆的总和 int under[maxn]; ///带表当前节点下有多少pile int n; void Init() { for(int i=0; i<=n; i++) { sum[i] = 1; fa[i] = i; } memset(under,0,sizeof(under)); } int Find(int u) { int tmp; if(u != fa[u]) { tmp = Find(fa[u]); under[u] += under[fa[u]]; fa[u] = tmp; } return fa[u]; } void Union(int x,int y) { int X,Y; X = Find(x); Y = Find(y); if (X!=Y) { under[X] = sum[Y]; //X是当前堆(集合)中最底部的,直接更新under[] sum[Y] += sum[X]; //直接更新Y这堆(集合)的高度(总共多少个Piles) fa[X] = Y; //合并 } } int main() { while(~scanf("%d",&n)) { Init(); char ch[10]; int u,v,w; while(n--) { scanf("%s",ch); if(ch[0] == ‘M‘) { scanf("%d %d",&u,&v); Union(u,v); } else { scanf("%d",&w); Find(w); printf("%d\n",under[w]); } } } return 0; }
原文:http://www.cnblogs.com/chenyang920/p/4359440.html