Cube Stacking
| Time Limit: 2000MS |
|
Memory Limit: 30000K |
| Total Submissions: 18858 |
|
Accepted: 6547 |
| 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
题意：n个方块（n<=30000）,p组操作（p<=100000）,操作有两种，M a b 将含有a的堆放在包含b的堆上，还有一种是 C a统计a下面有多少个方块
思路：带权并查集，一堆中最顶上的方块作为父节点，用dist[X] 统计X到父亲节点的距离，rank[fa[X]]表示团的大小，两者相减即为答案
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
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];
}
return fa[x];
}
int main(){
int n;
while(~scanf("%d",&n)){
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];
rank[faA] += rank[faB];
}
}else{
scanf("%d",&a);
int x = find(a);
printf("%d\n",rank[x]-dist[a]-1);
}
}
}
return 0;
}
POJ1988-Cube Stacking(带权并查集),布布扣,bubuko.com
POJ1988-Cube Stacking(带权并查集)
原文：http://blog.csdn.net/mowayao/article/details/38311599
