SRM530

	Level 1	Level 2	Level 3
Div 1	Level 1	Level 2	Level 3
Div 2	Level 1	Level 2	Level 3

Tutorials:

SRM530

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Division One - Level Three:

Solution

Source Code:

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Division One - Level Two:

Solution

Source Code:

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Division One - Level One:

Solution

The same with Div2-Lev2.

Source Code:

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Sivision Two - Level Three:

Solution

Source Code:

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Division Two - Level Two:

Solution

Brute-force. Even here I made a minor mistake, which assume the cutter will start with a dot, what a joke.

Source Code:

/**
 * @author antonio081014
 * @since Jan 19, 2012, 6:25:15 PM
 */

public class GogoXCake {
    public String solve(String[] cake, String[] cutter) {
        for (int i = 0; i + cutter.length <= cake.length; i++) {
            for (int j = 0; j + cutter[0].length() <= cake[i].length(); j++) {
                if (checkCutter(cake, cutter, i, j)) {
                    for (int x = 0; x < cutter.length; x++) {
                        for (int y = 0; y < cutter[x].length(); y++) {
                            if (cutter[x].charAt(y) == '.')
                                cake[i + x] = cake[i + x].substring(0, y + j)
                                        + "X"
                                        + cake[i + x].substring(y + j + 1);
                        }
                    }
                }
            }
        }
        return checkCake(cake) ? "YES" : "NO";
    }

    public void print(String[] cake) {
        for (int i = 0; i < cake.length; i++)
            System.out.println(cake[i]);
    }

    public boolean checkCake(String[] cake) {
        for (int i = 0; i < cake.length; i++)
            for (int j = 0; j < cake[i].length(); j++)
                if (cake[i].charAt(j) == '.')
                    return false;
        return true;
    }

    public boolean checkCutter(String[] cake, String[] cutter, int x, int y) {
        for (int i = 0; i < cutter.length; i++) {
            for (int j = 0; j < cutter[i].length(); j++) {
                if (cutter[i].charAt(j) == '.')
                    if (cake[i + x].charAt(j + y) != cutter[i].charAt(j)) {
                        return false;
                    }
            }
        }
        return true;
    }

    // <%:testing-code%>
}
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Division Two - Level One:

Solution

I used the brute-force, which go through all the permutations.

Source Code:

//Thursday, January 19, 2012 17:51 PST
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

class GogoXBallsAndBinsEasy {
public:
  int solve(vector <int>);
};

int GogoXBallsAndBinsEasy::solve(vector <int> T) {
  vector<int> v(T.begin(), T.end());
  sort(v.begin(), v.end());
  sort(T.begin(), T.end());
  int count = 0;
  do{
    int tmp = 0;
    for(int i=0; i<v.size(); i++){
      tmp += (v[i] - T[i]) >0? (v[i]-T[i]) : 0;
    }
    count = max(count, tmp>0? tmp:-tmp);
  }while(next_permutation(v.begin(), v.end()));
  return count;
}


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