Description
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
* 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
Sample Input
6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4
Sample Output
1 0 2
本思路简单，但是没有路径压缩和优化，因此虽然结果正确，但是会超时，到现在也没想出怎样进行路径压缩
就是用数组记录下父节点和子节点，从而形成一个没有树枝的树，但是这样最大的坏处就是在遍历时会消耗大量的时间，不可取。
#include<iostream>
#include <cstdio>
using namespace std;
int count ;
int son[30005],fa[30005];
int find(int w)
{
if(fa[w]==0)
return w;
else
{
count++;
return find(fa[w]);
}
}
int findson(int w)
{
if(son[w]==0)
return w;
else
findson(son[w]);
}
int main()
{
int n,i,x,y,w;
scanf("%d",&n);
memset(son,0,4*30005);
memset(fa,0,4*30005);
for(i=1;i<=n;i++)
{
count=0;
char ch;
getchar();
scanf("%c",&ch);
if(ch=='M')
{
scanf("%d%d",&x,&y);
x=find(x);
y=findson(y);
fa[x]=y;
son[y]=x;
}
else
{
scanf("%d",&w);
find(w);
printf("%d\n",count);
}
}
return 0;
}
思路稍做优化，依旧超时.........但是结果依旧是正确的
#include<iostream>
#include <cstdio>
using namespace std;
int son[30005],fa[30005],count[30005];
int findfa(int x,int y)
{
if(fa[x]==0)
{
count[x]=count[x]+count[y];
return x;
}
else
{
count[x]=count[x]+count[y];
return findfa(fa[x],y);
}
}
int findson(int w)
{
if(son[w]==0)
return w;
else
findson(son[w]);
}
void Count(int x,int cou )
{
count[x]=cou+count[x];
if(son[x]==0)
return ;
Count(son[x],cou);
}
int main()
{
int n,i,x,y,w;
scanf("%d",&n);
memset(son,0,4*30005);
memset(fa,0,4*30005);
for( i=0;i<30005;i++)
count[i]=1;
for(i=1;i<=n;i++)
{
char ch;
getchar();
scanf("%c",&ch);
if(ch=='M')
{
scanf("%d%d",&x,&y);
y=findson(y);
Count(x,count[y]);
x=findfa(x,y);
// cout<<"****************"<<x<<' '<<y<<endl;
fa[x]=y;
son[y]=x;
}
else
{
scanf("%d",&w);
printf("%d\n",count[w]-1) ;
}
}
return 0;
}
正解：
#include<cstdio>
#include<cstring>
const int maxn=100000+5;
int set[maxn],cnt[maxn],top[maxn]; //t[k]为k所在栈的栈底元素序号，top[k]为k所在栈的栈顶元素序号，cnt[k]为k到set[k]的元素个数
int set_find(int p) //计算元素p所在栈的栈底元素set[p]以及p到set[p]的元素个数，路径压缩
{
if(set[p]<0) //栈中只有一个元素p
return p;
if(set[set[p]]>=0) //的下方还有元素
{
int fa=set[p];
set[p]=set_find(fa);
cnt[p]=cnt[p]+cnt[fa]; //累加从fa到set[p]之间的元素个数,计算出每个元素到栈底的距离
}
return set[p];
}
void set_join(int x,int y) //将x所在的栈移动到y所在的栈的栈顶
{
x=set_find(x); //将x，y分别设置为相应栈的栈底元素的序号
y=set_find(y);i
set[x]=y; //将y设置为set[x]的下一元素，两个栈连接完成
set_find(top[y]); //计算原先y所在栈的栈顶元素到栈底元素之间的元素个数
cnt[x]=cnt[top[y]]+1; //计算新栈从x到栈底的元素个数
top[y]=top[x]; //更新y所在栈的栈顶元素为x原来所在栈的栈顶元素
}
int main()
{
int p;
scanf("%d",&p); //输入操作的数目
memset(set,-1,sizeof(set)); //将set中所有的元素都独自成栈
memset(cnt,0,sizeof(cnt)); //清空计数
for(int i=0; i<maxn; i++) //初始化每一个栈的栈顶元素
top[i]=i;
while(p--)
{
char s[5];
scanf("%s",s); //读入当前操作
if(s[0]=='M')
{
int x,y;
scanf("%d%d",&x,&y);
set_join(x,y); //移动栈
}
else
{
int x;
scanf("%d",&x);
set_find(x);
printf("%d\n",cnt[x]);
}
}
return 0;
}
