Cube Stacking（点击转到）
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 28145 | Accepted: 9879 | |
| 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
1.题目含义：
有N(N<=30,000)堆方块，开始每堆都是一个方块。方块编号1 – N. 有两种操作：
M x y ： 表示把方块x所在的堆，拿起来叠放到y所在的堆上。
C x : 问方块x下面有多少个方块。
操作最多有 P (P<=100,000)次。对每次C操作，输出结果。
2.牵扯到多组合并，多组查询，考虑用并查集。
2.1 并查集思路或者说模板就是：初始化——查找——合并 的过程，对应不同的题目，在后两个过程中会有相应的一些改变。
2.2 本题就多了在查找，合并过程中，更新under[i]和sum[i]的过程。（因为题目中，要求输出查询结果，也就是当前值的下面有多少个方块。所以定义了一个under[]数组保存，因为更新under[]数组的过程中，需要知道父节点为根的总结点数，故而定义sum[]数组）
3.
34合并到12上面的时候，under[3]=sum[1] sum[1]=sum[1]+sum[3]，然后一直重新赋值。
4.一开始超时，超时原因用的cin读入。
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 31000
int par[maxn];
int sum[maxn]; //表示砖块所在树的总结点数
int under[maxn];//under[i]表示i下面有多少块
int p;
void init()
{
for(int i=0;i<maxn;i++)
{
par[i]=i;
sum[i]=1;
under[i]=0;
}
}
int getRoot(int x)
{
if(par[x]!=x)
{
int t=getRoot(par[x]);
under[x]=under[x]+under[par[x]];//见图片
par[x]=t;
}
return par[x];
}
void Merge(int x,int y)
{
int fx=getRoot(x);
int fy=getRoot(y);
if(fx!=fy)
under[fy]=sum[fx],sum[fx]=sum[fx]+sum[fy],par[fy]=fx;//见图片
}
int main()
{
char c;
int x;
int y;
init();
scanf("%d",&p);
for(int i=0;i<p;i++)
{
char s[20];
int a,b;
scanf("%s",s);
if( s[0] == 'M')
{
scanf("%d%d",&a,&b);
Merge(b,a);
}
else
{
scanf("%d",&a);
getRoot(a);
printf("%d\n",under[a]);
}
}
return 0;
}5.getchar()读走\n
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 31000
int par[maxn];
int sum[maxn]; //表示砖块所在树的总结点数
int under[maxn];//under[i]表示i下面有多少块
int p;
void init()
{
for(int i=0;i<maxn;i++)
{
par[i]=i;
sum[i]=1;
under[i]=0;
}
}
int getRoot(int x)
{
if(par[x]!=x)
{
int t=getRoot(par[x]);
under[x]=under[x]+under[par[x]];//见图片
par[x]=t;
}
return par[x];
}
void Merge(int x,int y)
{
int fx=getRoot(x);
int fy=getRoot(y);
if(fx!=fy)
under[fy]=sum[fx],sum[fx]=sum[fx]+sum[fy],par[fy]=fx;//见图片
}
int main()
{
char c;
int x;
int y;
init();
scanf("%d",&p);
for(int i=0;i<p;i++)
{
char s;
int a,b;
getchar();
scanf("%c",&s);
if( s == 'M')
{
scanf("%d%d",&a,&b);
Merge(b,a);
}
else
{
scanf("%d",&a);
getRoot(a);
printf("%d\n",under[a]);
}
}
return 0;
}