SRM233

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

Tutorials:

Division One - Level Three:

Solution

Source Code:

Division One - Level Two:

Solution

Breadth First Search (BFS).
Problem hints:

• Words can be considered as states. There are at most 26^4 different words composed of 4 letters (thus a linear complexity will run in allowed time).
• There are some ways to pass from one state to another.
• The cost of passing from a state to another is always 1 (i.e. a single button click).
• You need to find the minimum number of steps required to reach the end state from start state.

Source Code:

//Thu May 19 11:12:55 CDT 2011
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#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 <cctype>
#include <string>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

int dx[] = { 0, 0, 1, -1, 0, 0, 0, 0 };
int dy[] = { 0, 0, 0, 0, 1, -1, 0, 0 };
int dz[] = { 0, 0, 0, 0, 0, 0, 1, -1 };
int dw[] = { 1, -1, 0, 0, 0, 0, 0, 0 };

class State {
public:
    int a;
    int b;
    int c;
    int d;
    State(string s) {
        this->a = s[0] - 'a';
        this->b = s[1] - 'a';
        this->c = s[2] - 'a';
        this->d = s[3] - 'a';
    }
    State(int aa, int bb, int cc, int dd) {
        this->a = aa;
        this->b = bb;
        this->c = cc;
        this->d = dd;
    }
    bool equals(State s) {
        return this->a == s.a && this->b == s.b && this->c == s.c && this->d
                == s.d;
    }
};

class SmartWordToy {

private:
    bool forbiden[26][26][26][26];
    int cost[26][26][26][26];
public:
    int minPresses(string start, string finish, vector<string> forbid) {
        init(forbid);
        //      if (forbiden[start[0] - 'a'][start[1] - 'a'][start[2] - 'a'][start[3]
        //              - 'a'])
        //          return -1;
        State st(start);
        cost[start[0] - 'a'][start[1] - 'a'][start[2] - 'a'][start[3] - 'a']
                = 0;
        queue<State> q;
        q.push(st);
        while (!q.empty()) {
            int a = q.front().a;
            int b = q.front().b;
            int c = q.front().c;
            int d = q.front().d;
            int dist = cost[a][b][c][d];
            q.pop();
            for (int i = 0; i < 8; i++) {
                int x = (a + dx[i] + 26) % 26;
                int y = (b + dy[i] + 26) % 26;
                int z = (c + dz[i] + 26) % 26;
                int w = (d + dw[i] + 26) % 26;
                State tmpST(x, y, z, w);
                if (forbiden[x][y][z][w] || cost[x][y][z][w] != -1)
                    continue;
                cost[x][y][z][w] = dist + 1;
                q.push(tmpST);
            }
        }
        return cost[finish[0] - 'a'][finish[1] - 'a'][finish[2] - 'a'][finish[3]
                - 'a'];
    }

    void init(vector<string> f) {
        for (int i = 0; i < 26; i++) {
            for (int j = 0; j < 26; j++) {
                for (int p = 0; p < 26; p++) {
                    for (int q = 0; q < 26; q++) {
                        forbiden[i][j][p][q] = false;
                        cost[i][j][p][q] = -1;
                    }
                }
            }
        }
        for (int i = 0; i < (int) f.size(); i++) {
            vector<string> tmp = split(f[i]);
            for (int a = 0; a < (int) tmp[0].size(); a++) {
                for (int b = 0; b < (int) tmp[1].size(); b++) {
                    for (int c = 0; c < (int) tmp[2].size(); c++) {
                        for (int d = 0; d < (int) tmp[3].size(); d++) {
                            forbiden[tmp[0][a] - 'a'][tmp[1][b] - 'a'][tmp[2][c]
                                    - 'a'][tmp[3][d] - 'a'] = true;

                        }
                    }
                }
            }
        }
    }

    vector<string> split(string s) {
        vector<string> ret;
        stringstream ss(s);
        while (ss >> s) {
            ret.push_back(s);
        }
        return ret;
    }
};

Division One - Level One:

Solution

Same with Div2, Level2.

Source Code:

Division Two - Level Three:

Solution

Source Code:

Division Two - Level Two:

Solution

All cuts / C(N,2).

Source Code:

//SRM233Div2_500;
//SRM233DIV1_250;
//2009/10/23 16:28:28
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class PipeCuts
{
public:
    double probability(vector <int> weldLocations, int L)
    {
        int count = 0;
        sort(weldLocations.begin(), weldLocations.end());
        for(int i=0; i<weldLocations.size(); i++)
        {
            for(int j=i+1; j<weldLocations.size(); j++)
                if((weldLocations[j]-weldLocations[i]>L) || weldLocations[i]>L || (100-weldLocations[j])>L)
                    count ++;
        }
        return 2.0 * count / ((weldLocations.size()-1) * weldLocations.size());
    }
};

Division Two - Level One:

Solution

Source Code:

//2009/08/15 18:43:18
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <sstream>
#include <algorithm>

using namespace std;

class JustifyText
{
public:
    vector <string> format(vector <string> text)
    {
        int sz = 0;
        vector<string> ret;
        for(int i=0; i<text.size(); i++) sz = max((int)text[i].size(), sz);
        for(int i=0; i<text.size(); i++)
        {
            string s(sz-text[i].size(), ' ');
            ret.push_back(s+text[i]);
        }
        return ret;
    }
};