minimum cost path by Dynamic programming

#include<iostream>
using namespace std;

//first method
int MinCost(int C[][3],int r,int c){
    int MC[r+1][c+1]={0,0};
    MC[0][0]=C[0][0];
    for(int i=0;i<=r;i++){
      for(int j=0;j<=c;j++){
         if(MC[i][j]==0){
                int first,second,third;
                first=(i-1>=0)? MC[i-1][j]:INT_MAX;
                second=(j-1>=0)? MC[i][j-1]:INT_MAX;
                third=(i-1>=0 && j-1 >=0)? MC[i-1][j-1]:INT_MAX;
                MC[i][j]=C[i][j]+min(first,min(second,third));
               }
          }
       }
    return MC[r][c];
    }

  //second method from geeks for geeks
int MinCost(int C[][3],int r,int c){
  int MC[r+1][c+1];
  MC[0][0]=C[0][0];
  for(int i=1;i<=r;i++)
    MC[i][0]=MC[i-1][0]+C[i][0];
  for(int i=1;i<=c;i++)
    MC[0][i]=MC[0][i-1]+C[0][i];
  for(int i=1;i<=r;i++){
    for(int j=1;j<=c;j++){
      MC[i][j]=min(MC[i-1][j-1], min(MC[i-1][j],MC[i][j-1]))+C[i][j];
      }
    }
  return MC[r][c];
  }

int main(){
int r=3,c=3;
int C[r][3]={ {1,2,3},
              {4,8,2},
              {1,5,3}
               };
cout<<MinCost(C,2,2);
  return 0;
  }