CareerCup

/*
Solution:
- looks like from hackerrank/topcoder, sure it was an interview?
- if you are at position (x,y) you can go either
  to (x+1,y) or (x,y+1) both has an assosiated cost depending on
  the direction you will be looking (0,1) where 0 means facing down
  and 1 facing to the right:
  - if you come from the left field, you will look right and only cost
    is move cost
  - if you come from top you will look down only cost is move
    cost
  - if facing right is cheaper when coming from top + 
    turning, use this cost
  - same from coming from left + turning for facing down
  so, there are two cost per field, one is reaching it facing
  down and the other is reaching it and facing right.
  ...
- The O(n*m) iterative solution is as follows:
  - c(d,x,y) denotes the cost at field x,y facing d: (d=0: down, d=1: 
    right), 0 <= x < M, 0 <= y < N	
	for(int y = 0; y < M; y++)
		for(int x = 0; x < N; x++)
			int c0 = MAX;
			int c1 = MAX;
			if(x > 0) 
			{
				c1 = c[1][y][x-1] + move_cost[y][x];
				c0 = c1 + turn_cost[y][x];
			}
			if(y > 0)
			{
				c0 = min(c0, c[0][y-1][x] + move_cost[y][x])
				c1 = min(c1, c0 + turn_cost[y][x])
			}
			if(x == 0 && y == 0) 
			{ // start case
				c0 = 0;
				c1 = 0;
			}
			c[0][y][x] = c0;
			c[1][y][x] = c1;
	
	return min(c[0][N-1][M-1], c[1][N-1][M-1]);

	complete solution in c++11:
*/

#include <iostream>
#include <string>
#include <vector>
#include <utility>
#include <cassert>
#include <limits>

using namespace std;

int findMinCostToTraverseWithTurn(const vector<vector<int>>& move_cost, const vector<vector<int>>& turn_cost) 
{
	assert(move_cost.size()  > 0 && turn_cost.size() == move_cost.size());
	assert(move_cost[0].size()  > 0 && turn_cost[0].size() == move_cost[0].size());
	int n = move_cost.size();
	int m = move_cost[0].size();

	vector<vector<vector<int>>> c(2, 
		vector<vector<int>>(n,
		vector<int>(m, numeric_limits<int>::max())));

	// start case
	c[0][0][0] = 0;
	c[1][0][0] = 0;

	for(int y = 0; y < n; y++)
	{
		for(int x = 0; x < m; x++)
		{
			if(x > 0) 
			{
				c[1][y][x] = c[1][y][x - 1] + move_cost[y][x];
				c[0][y][x] = c[1][y][x] + turn_cost[y][x];
			}
			if(y > 0)
			{
				c[0][y][x] = min(c[0][y][x], c[0][y - 1][x] + move_cost[y][x]);
				c[1][y][x] = min(c[1][y][x], c[0][y][x] + turn_cost[y][x]);
			}
			// cout << "(" << c[0][y][x] << "," << c[1][y][x] << ") ";
		}
		// cout << endl;
	}
	return min(c[0][n - 1][m - 1], c[1][n - 1][m - 1]);
}


int main()
{
	vector<vector<int>> move_cost { 
		{0,1,2},
		{1,2,1},
		{2,1,2}
	};
	vector<vector<int>> turn_cost { 
		{0,2,3},
		{1,0,1},
		{0,0,1}
	};
	cout << findMinCostToTraverseWithTurn(move_cost, turn_cost);
}

#include<bits/stdc++.h>
using namespace std;
int arr[1005][1005];
int mem[1005][1005][2];
int turn[1005][1005];
int main()
{
    freopen("out.txt","w",stdout);
     freopen("in.txt","r",stdin); 
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            {
                cin>>arr[i][j];
        
    }
    for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
            {
                int x;
                cin>>turn[i][j];
        }
    turn[n-1][m-1]=0;
    mem[0][0][0]=arr[0][0];
    mem[0][0][1]=arr[0][0];
    for(int i=1;i<m;i++)
        {
        mem[0][i][0]=arr[0][i]+mem[0][i-1][0];
        mem[0][i][1]=arr[0][i]+mem[0][i-1][0]+turn[0][i];
        
    }
    for(int i=1;i<n;i++)
        {
        mem[i][0][0]=arr[i][0]+mem[i-1][0][1]+turn[i][0];
        mem[i][0][1]=arr[i][0]+mem[i-1][0][1];
    }
    for(int i=1;i<n;i++)
        for(int j=1;j<m;j++)
            {
                mem[i][j][0]=arr[i][j]+min(mem[i-1][j][1]+turn[i][j],mem[i][j-1][0]);
                mem[i][j][1]=arr[i][j]+min(mem[i-1][j][1],mem[i][j-1][0]+turn[i][j]);
            }
    cout<<mem[n-1][m-1][1];
    return 0;
}

@Rohit. any example inputs?

#include<bits/stdc++.h>
using namespace std;
int arr[1005][1005];
int mem[1005][1005][2];
int turn[1005][1005];
int main()
{
freopen("out.txt","w",stdout);
freopen("in.txt","r",stdin);
int n,m;
cin>>n>>m;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
cin>>arr[i][j];

}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
int x;
cin>>turn[i][j];
}
turn[n-1][m-1]=0;
mem[0][0][0]=arr[0][0];
mem[0][0][1]=arr[0][0];
for(int i=1;i<m;i++)
{
mem[0][i][0]=arr[0][i]+mem[0][i-1][0];
mem[0][i][1]=arr[0][i]+mem[0][i-1][0]+turn[0][i];

}
for(int i=1;i<n;i++)
{
mem[i][0][0]=arr[i][0]+mem[i-1][0][1]+turn[i][0];
mem[i][0][1]=arr[i][0]+mem[i-1][0][1];
}
for(int i=1;i<n;i++)
for(int j=1;j<m;j++)
{
mem[i][j][0]=arr[i][j]+min(mem[i-1][j][1]+turn[i][j],mem[i][j-1][0]);
mem[i][j][1]=arr[i][j]+min(mem[i-1][j][1],mem[i][j-1][0]+turn[i][j]);
}
cout<<mem[n-1][m-1][1];
return 0;
}