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  1. //
  2. //  돌_그룹_12886.cpp
  3. //  BFS
  4. //
  5. //  Created by jiho park on 2020/01/25.
  6. //  Copyright © 2020 jiho park. All rights reserved.
  7. //
  8.  
  9. #include <iostream>
  10. #include <queue>
  11. #include <tuple>
  12.  
  13. using namespace std;
  14.  
  15. bool check[501][501][501];
  16. int a[3];
  17.  
  18. pair<int,int> cal(int x, int y){
  19.     if(x<y){
  20.         y=y-x;
  21.         x=x+x;
  22.     }else{
  23.         x=x-y;
  24.         y=y+y;
  25.     }
  26.     return make_pair(x,y);
  27. }
  28.  
  29. int main(){
  30.     cin >> a[0] >> a[1] >> a[2];
  31.     if(a[0]==a[1] && a[1]==a[2]){
  32.         cout << "1";
  33.         return 0;
  34.     }
  35.  
  36.     queue<tuple<int,int,int>> q;
  37.     check[a[0]][a[1]][a[2]]=true;
  38.     q.emplace(a[0],a[1],a[2]);
  39.  
  40.     while(!q.empty()){
  41.         int x[3];
  42.         tie(x[0],x[1],x[2]) = q.front();
  43.         q.pop();
  44.         for(int i=0; i<2; i++){
  45.             for(int j=i+1; j<3; j++){
  46.                 if(x[i] != x[j]){
  47.                     int nx, ny;
  48.                     tie(nx,ny) = cal(x[i], x[j]);
  49.                     if(i==0 && j==1 && !check[nx][ny][x[2]]){
  50.                         if(nx==ny && ny==x[2]){
  51.                             cout << "1";
  52.                             return 0;
  53.                         }
  54.                         q.emplace(nx,ny,x[2]);
  55.                         check[nx][ny][x[2]]=true;
  56.                     }
  57.                     else if(i==0 && j==2 && !check[nx][x[1]][ny]){
  58.                         if(nx==x[1] && x[1]==ny){
  59.                             cout << "1";
  60.                             return 0;
  61.                         }
  62.                         q.emplace(nx,x[1],ny);
  63.                         check[nx][x[1]][ny]=true;
  64.                     }
  65.                     else if(i==1 && j==2 && !check[x[0]][nx][ny]){
  66.                         if(x[0]==nx && nx==ny){
  67.                             cout << "1";
  68.                             return 0;
  69.                         }
  70.                         q.emplace(x[0],nx,ny);
  71.                         check[x[0]][nx][ny]=true;
  72.                     }
  73.                 }
  74.             }
  75.         }
  76.     }
  77.  
  78.     cout << "0";
  79.  
  80.     return 0;
  81. }
  82.  
Success #stdin #stdout 0s 4296KB
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https://ideone.com/8704nS
language:
C++14 (gcc 8.3)
created:
4 years ago
visibility:
public

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