hdu 2818 Building Block
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2818
分析:带权并查集,给出两种操作,分别是并和查。 不过合并方式并不相同,要求将x中的元素合并到y中。
/*Building Block Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3484 Accepted Submission(s): 1047 Problem Description John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation: M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command. C X : Count the number of blocks under block X You are request to find out the output for each C operation. Input The first line contains integer P. Then P lines follow, each of which contain an operation describe above. Output Output the count for each C operations in one line. Sample Input 6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4 Sample Output 1 0 2 Source 2009 Multi-University Training Contest 1 - Host by TJU */ #include <cstdio> #include <cstring> const int maxn = 30000 + 100; /* markset()用以初始化 findset()查询 unin() 合并 */ struct lset{ int p[maxn], rank[maxn], sum[maxn], sz; void makeset() { memset(rank, 0, sizeof(rank)); memset(sum, 0, sizeof(sum)); for(int i = 0; i < maxn; i++){ p[i] = i; } } int findset(int x) { if(x != p[x]){ int fa = p[x]; p[x] = findset(fa); rank[x] += rank[fa]; } return p[x]; } void unin(int x, int y) { int fx = findset(x), fy = findset(y); if(fx != fy){ p[fx] = fy; rank[fx] += sum[fy] + 1; sum[fy] += sum[fx] + 1; } } void compress() { for(int i = 0; i < sz; i++) findset(i); } }; int main() { int p, x, y; char c; lset B; while(~scanf("%d%*c", &p)){ B.makeset(); for(int i = 0; i < p; i++){ c = getchar(); if(c == 'M'){ scanf("%d%d%*C", &x, &y); B.unin(x, y); } else{ scanf("%d%*C", &x); B.findset(x); printf("%d\n", B.rank[x]); } } } return 0; }