|
Cube Stacking
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 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 |
|||||||||
题目大意：
有N个立方体和N个格子，1~N编号，一開始i立方体在i号格子上，每一个格子刚好1个立方体。如今m组操作，M a b表示将a号立方体所在的格子的所有立方体放在b号立方体所在的格子的所有立方体上面。C x表示询问x号立方体以下的立方体的个数。
解题思路：
在并查集的基础上。仅仅须要知道x到父亲的距离以及父亲究竟的距离就知道x究竟的距离。
解题代码：
#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=31000;
int father[maxn],cnt[maxn],dis[maxn];
int find(int x){
if(father[x]!=x){
int tmp=father[x];
father[x]=find(father[x]);
dis[x]+=dis[tmp];
}
return father[x];
}
void combine(int x,int y){
father[x]=y;
dis[x]+=cnt[y];
cnt[y]+=cnt[x];
}
int main(){
int m;
scanf("%d",&m);
for(int i=0;i<maxn;i++){
father[i]=i;
cnt[i]=1;
dis[i]=0;
}
while(m-- >0){
char ch;
cin>>ch;
if(ch=='M'){
int a,b;
scanf("%d%d",&a,&b);
if(find(a)!=find(b)) combine(find(a),find(b));
}else{
int x;
scanf("%d",&x);
find(x);
printf("%d\n",dis[x]);
}
}
return 0;
}
版权声明：本文博客原创文章。博客，未经同意，不得转载。