Problem:
Pile Up!
There are cubes of the same size and a simple robot named Masato. Initially, 
all cubes are on the floor. Masato can be instructed to pick up a cube and put 
it on another cube, to make piles of cubes. Each instruction is of the form 
`pick up cube
A
and put it on cube
B
(or on the floor).'
When he is to pick up a cube, he does so after taking off all the cubes on 
and above it in the same pile (if any) onto the floor. In contrast, when he is 
to put a cube on another, he puts the former on top of the pile including the 
latter without taking any cubes off.
When he is instructed to put a cube on another cube and both cubes are 
already in the same pile, there are two cases. If the former cube is stacked 
somewhere below the latter, he put off all the cube above it onto the floor. 
Afterwards he puts the former cube on the latter cube. If the former cube is 
stacked somewhere above the latter, he just ignores the instruction.
When he is instructed to put a cube on the floor, there are also two cases. 
If the cube is already on the floor (including the case that the cube is stacked 
in the bottom of some pile), he just ignores the instruction. Otherwise, he puts 
off all the cube on and up above the pile (if any) onto the floor before moving 
the cube onto the floor.
He also ignores the instruction when told to put a cube on top of itself 
(this is impossible).
Given the number of the cubes and a series of instructions, simulate actions 
of Masato and calculate the heights of piles when Masato has finished his job.
Input
The input consists of a series of data sets. One data set contains the number 
of cubes as its first line, and a series of instructions, each described in a 
separate line. The number of cubes does not exceed 100. Each instruction 
consists of two numbers; the former number indicates which cube to pick up, and 
the latter number indicates on which cube to put it. The end of the instructions 
is marked by two zeros.
The end of the input is marked by the line containing a single zero.
One data set has the following form:
m
I
1
J
1
I
2
J
2
...
I
n
J
n
0 0
Each cube is identified by its number (1 through
m
).
I
k
indicates which cube to pick up and
J
k
indicates on which cube to put it. The latter may be 
zero, instructing to put the cube on the floor.
Output
Output the height of each pile (the number of cubes in the pile) in ascending 
order, separated by newlines. A single cube by itself also counts as "a pile." 
End of output for one data set should be marked by a line containing the 
character sequence `end' by itself.
Sample Input
3
1 3
2 0
0 0
4
4 1
3 1
1 2
0 0
5
2 1
3 1
4 1
3 2
1 1
0 0
0
Output for the Sample Input
1
2
end
1
1
2
end
1
4
end

Submitted solutions:

code_snippet: #include "bits/stdc++.h"
#include<unordered_map>
#include<unordered_set>
#pragma warning(disable:4996)
using namespace std;
using ld = long double;
const ld eps = 1e-9;

//// < "d:\d_download\visual studio 2015\projects\programing_contest_c++\debug\a.txt" > "d:\d_download\visual studio 2015\projects\programing_contest_c++\debug\b.txt"


int main() {
	while (1) {
		int M; cin >> M;
		if (!M)break;
		vector<vector<int>>piles(M);
		for (int i = 0; i < piles.size(); ++i) {
			piles[i] = vector<int>(1, i + 1);
		}
		while (1) {
			int a, b; cin >> a >> b;
			if (!a)break;
			pair<int, int>ap;
			for (int i = 0; i < piles.size(); ++i) {
				auto pile(piles[i]);
				auto it = find(pile.begin(), pile.end(), a);
				if (it != pile.end()) {
					ap = make_pair(i, it - pile.begin());
				}
			}
			if (!b) {
				if (ap.second == 0)continue;
				auto&apile(piles[ap.first]);
				while (apile.back() != a) {
					int n = apile.back();

					apile.pop_back();
					if (n != a) {

						piles.push_back(vector<int>(1, n));
					}
				}
				apile.pop_back();
				piles.push_back(vector<int>(1, a));
			}
			else {
				pair<int, int> bp;
				
				for (int i = 0; i < piles.size(); ++i) {
					auto pile(piles[i]);
					auto jt = find(pile.begin(), pile.end(), b);
					if (jt != pile.end()) {
						bp = make_pair(i, jt - pile.begin());
					}
				}

				if (ap.first == bp.first) {
					if (ap.second >= bp.second) {
						continue;
					}
					else {
						auto& apile(piles[ap.first]);
						while (apile.back() != a) {
							int n = apile.back();

							apile.pop_back();
							if (n != b) {

								piles.push_back(vector<int>(1, n));
							}
						}
						apile.pop_back();
						piles.push_back(vector<int>{b, a});
					}
				}
				else {
					auto&apile(piles[ap.first]);
					auto&bpile(piles[bp.first]);
					while (apile.back() != a) {
						int n = apile.back();

						apile.pop_back();
						if (n != a) {

							piles.push_back(vector<int>(1, n));
						}
					}
					apile.pop_back();
					bpile.push_back(a);
				}
			}
			
			{

				vector<vector<int>>nextpiles;
				for (auto&pile : piles) {
					if (!pile.empty())nextpiles.push_back(pile);
				}
				piles = nextpiles;
			}
		}
		vector<int>nums;
		for (auto pile : piles) {
			nums.push_back(pile.size());
		}
		sort(nums.begin(), nums.end());
		for (auto n : nums)cout << n << endl;
		cout << "end" << endl;
	}
	return 0;
}
accuracy: 1
time_ms: 10
memory_kb: 3048
final_rank: 2

code_snippet: #include<bits/stdc++.h>
#define N 105
using namespace std;
int m,I,J,A[N][N];
vector<int> ans;

void func1(int a,int b){
  int ay,ax,by,bx;
  for(int i=0;i<m;i++)
    for(int j=0;j<m;j++){
      if(a==A[i][j])ay=i,ax=j;
      if(b==A[i][j])by=i,bx=j;
    }
  A[ay][ax]=0;
  for(int i=ax+1;i<m;i++){
    if(!A[ay][i])break;
    for(int j=0;j<m;j++)
      if(!A[j][0]){
	A[j][0]=A[ay][i];
	A[ay][i]=0;
	break;
      }
  }

  for(int i=bx+1;i<m;i++)
    if(!A[by][i]){
      A[by][i]=a;
      break;
    }
}

void func2(int a,int b){
  int ay,ax,bx;
  for(int i=0;i<m;i++)
    for(int j=0;j<m;j++){
      if(a==A[i][j])ay=i,ax=j;
      if(b==A[i][j])bx=j;
    }
  if(ax>=bx)return;

  for(int i=ax;i<m&&A[ay][i];i++){
    for(int j=0;j<m;j++)
      if(!A[j][0]){
	A[j][0]=A[ay][i];
	A[ay][i]=0;
	break;
      }
  }
  func1(a,b);
}


void func3(int a){
  int ay,ax;
  for(int i=0;i<m;i++)
    for(int j=0;j<m;j++)
      if(a==A[i][j])ay=i,ax=j;

  if(!ax)return;
  
  for(int i=ax;i<m;i++){
    if(!A[ay][i])break;
    for(int j=0;j<m;j++)
      if(!A[j][0]){
	A[j][0]=A[ay][i];
	A[ay][i]=0;
	break;
      }
  }
}


bool check(int a,int b){
  int ay,by;
  for(int i=0;i<m;i++){
    for(int j=0;j<m;j++){
      if(a==A[i][j])ay=i;
      if(b==A[i][j])by=i;
    }
  }
  return ay==by;
}

int main(){

  while(1){
    cin>>m;
    if(!m)break;
    memset(A,0,sizeof(A));
    for(int i=0;i<m;i++)A[i][0]=i+1;

    while(1){
      cin>>I>>J;
      if(!I&&!J)break;
      if(I&&!J)func3(I);
      else if(!check(I,J))func1(I,J);
      else func2(I,J);
    }

    for(int i=0;i<m;i++){
      int cnt=0;
      for(int j=0;j<m;j++){
	if(!A[i][j])break;
	cnt++;
      }
      if(cnt)ans.push_back(cnt);
    }
    sort(ans.begin(),ans.end());
    for(int i=0;i<(int)ans.size();i++)cout<<ans[i]<<endl;
    cout<<"end"<<endl;
    ans.clear();
  }
  return 0;
}
accuracy: 1
time_ms: 50
memory_kb: 1264
final_rank: 3

code_snippet: #include <iostream>
#include <algorithm>
#include <cassert>
#include <cctype>
#include <complex>
#include <cstdio>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)

int m,I,J,p[10000][100],s[10000],P;
pair<int,int> pos[101];

void destroy(int pI,int ppI,pair<int,int> *pos){
	p[pI][ppI]=0;
	for(int i=ppI+1;i<s[pI];i++){
		p[P][0]=p[pI][i];
		s[P]=1;
		pos[p[P][0]]=make_pair(P++,0);
		p[pI][i]=0;
	}
	s[pI]=ppI;
}

int main(){
	while(cin>>m&&m){
		rep(i,10000)fill(p[i],p[i]+100,0);
		fill(s,s+10000,0);
		rep(i,m){p[i][0]=i+1;s[i]=1;pos[i+1]=make_pair(i,0);}
		P=m;
		while(cin>>I>>J&&I){
			if(I==J)continue;
			int pI=pos[I].first,ppI=pos[I].second,pJ=pos[J].first,ppJ=pos[J].second;
			if(!J){
				if(ppI){
					destroy(pI,ppI,pos);
					p[P][0]=I;
					s[P]=1;
					pos[I]=make_pair(P++,0);
				}
			}
			else if(pI==pJ){
				if(ppI>ppJ)continue;
				else{
					int pppJ=P+ppJ-ppI-1;
					destroy(pI,ppI,pos);
					p[pppJ][1]=I;
					s[pppJ]++;
					pos[I]=make_pair(pppJ,1);
				}
			}
			else{
				destroy(pI,ppI,pos);
				p[pJ][s[pJ]]=I;
				pos[I]=make_pair(pJ,s[pJ]++);
			}
			//cout<<I<<"->"<<J<<endl;
			//rep(i,m)cout<<i+1<<":"<<pos[i+1].first<<","<<pos[i+1].second<<(i<m-1?" ":"\n");
			//int sum=0;rep(i,P)sum+=s[i];cout<<sum<<endl;
			//rep(i,P){if(s[i])cout<<i<<":";rep(j,s[i]){cout<<p[i][j]<<(j<s[i]-1?" ":"");}if(s[i])cout<<endl;}cout<<endl;
		}
		sort(s,s+10000);
		rep(i,10000)if(s[i])cout<<s[i]<<(i<9999?"\n":"\nend\n");
	}
}
accuracy: 1
time_ms: 10
memory_kb: 0
final_rank: 1