poj1988（并查集）Cube Stacking

Cube Stacking Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 24968 Accepted: 8740 Case Time Limit: 1000MS Description Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations:  moves and counts.  * In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.  * In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value.  Write a program that can verify the results of the game.  Input * Line 1: A single integer, P  * Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X.  Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself.  Output Print the output from each of the count operations in the same order as the input file.  Sample Input 6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4 Sample Output 1 0 2 Source

USACO 2004 U S Open

题意： 有一些碟子，有两种操作，一是把x所在堆的所有碟子都叠到y所在的碟子上，二是询问x所下面有多少个碟子。 思路： 这是一道带权的并查集，跟之前做的差不多。设两个权值cnt[x]代表以x为根结点的堆碟子的个数，dist[x]表示x这碟子到根结点的距离。求x下面有多少碟子就等于cnt[pa[x]] - dist[x] - 1;

#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int M = 30005; int pa[M],cnt[M],dist[M]; void Init() { for (int i = 1;i <= M;i ++) pa[i] = i,cnt[i] = 1,dist[i] = 0; } int findset (int x) { if (x != pa[x]) { int root = findset (pa[x]); dist[x] += dist[pa[x]]; return pa[x] = root; } else return x; } void Unin(int a,int b) { int af = findset (a); int bf = findset (b); pa[bf] = af; dist[bf] = cnt[af]; cnt[af] += cnt[bf]; } int main () { int m,x,y; char op[3]; while (~scanf ("%d",&m)) { Init(); while (m --) { scanf ("%s",op); if (op[0] == 'M') { scanf ("%d%d",&x,&y); Unin(x,y); } else { scanf ("%d",&x); int af = findset (x); printf ("%d\n",cnt[af]-dist[x]-1); } } } return 0; }