poj 1079 Cube Stacking(并查集)
Cube Stacking
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 15705 | Accepted: 5331 | |
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:记录当前点集合的所有元素个数和当前点到根的距离,答案就是二者之差
#include<iostream> #include<cstring> using namespace std; const int mm=100010; class node { public:int r,u,dis; }g[mm]; int m; void dataset() { for(int i=0;i<mm;i++) g[i].r=i,g[i].u=0,g[i].dis=1; } int look(int z) { if(z^g[z].r)///压缩 { int p=g[z].r; g[z].r=look(g[z].r); g[z].u+=g[p].u; } return g[z].r; } void uni(int a,int b) { int x=look(a); int y=look(b); g[y].r=x; g[y].u=g[x].dis; g[x].dis+=g[y].dis; } int main() { char s;int a,b; while(cin>>m) { dataset(); while(m--) { cin>>s>>a; if(s=='C') { b=look(a);cout<<g[b].dis-g[a].u-1<<"\n"; } else {cin>>b;uni(a,b);} } } }
The article write by nealgavin