Cube Stacking
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 25647 | Accepted: 8975 | |
| 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.
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.
* 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
思路：
加权并查集
题意： 说是有n块砖，编号从1到n,有两种操作，第一是把含有x编号的那一堆砖放到含有编号y的那一堆砖的上面，第二是查询编号为x的砖的下面有多少块砖。用count[x]表示下面有多少块砖。
现在需要把两堆砖合并，显然要用上并查集，可是普通的合并之后如何知道x的下面有多少块砖呢，思考合并的过程，对于一堆砖，移动到另一堆砖上时，上面那一堆上每块砖的count[i]应该加上下面一堆砖的数量，这个操作对于上面一堆砖的根来说是简单的，我使用uset[i]表示连通分量，舒适化时所有的uset[i]为-1,负数代表这个节点为根，1代表这个连通分量的节点总数为1,以样例为例，首先将1放到6上面，即将6合并到1所在的连通分量中，合并的过程中我们知道两个信息，第一是当前连通分量6->1的节点数量为2,6距离1的距离为1,同理，将2放到4上面，这个连通分量节点个数为2,,4到2的距离为1,最后，我们将包含6的这个连通分量合并到包含2的这个连通分量中，此时连通分量数为4，曾经的6->1连通分量的根距离合并后的连通分量的根的距离为2,就是4->2的连通分量的节点数
说了半天有什么用处呢，经过上面这个过程，我们知道了每一个节点到它第一次被合并时的那个根节点的距离，6->1的距离为1,1到4的距离为2,2到4的距离为1,这样我们在查询6的下面有多少块砖时，直接用4(连通分量节点数)-（1+2）（6到根节点的距离）-1
上代码:
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int maxn = 30000+10; int fa[maxn]; int rank[maxn]; int dist[maxn]; void init() { for(int i=0; i<maxn; i++) { fa[i]=i; rank[i] = 1; dist[i] = 0; } } int find(int x) { if(x != fa[x]) { int t = fa[x]; fa[x] = find(fa[x]); dist[x] += dist[t];//dis[t]表示t，即x曾经的根节点距离根的距离，x到根的距离为x到t的距离加上t到根的距离 } return fa[x]; } int main() { int n; while(scanf("%d",&n)==1) { init(); char op; while(n--) { cin>>op; int a,b; if(op=='M') { scanf("%d%d",&a,&b); int faA = find(a); int faB = find(b); if(faA != faB) { fa[faB]=faA; dist[faB]=rank[faA];//下面的砖的根距离如今的连通分量的根的距离 rank[faA]+=rank[faB];//节点相加 } } else { scanf("%d",&a); int x = find(a); printf("%d ",rank[x]-dist[a]-1);//连通分量节点数-根节点的距离-1 } } } return 0; }
自己选的路，跪着也要走完！！！