Here you will find solutions of many problems on spoj. If you want solution of some problem which is not listed in blog or have doubt regarding any spoj problem (which i have solved) or any programming concept (data structure) you can mail me @ raj.nishant360@gmail.com
And my humble request to you all that don't copy the code only try to understand the logic and algorithm behind the code. I have started this because if you tried as hard as you can and still can't find any solution to the problem then you can refer to this.
You can read my answer how to start competitive programming CLICK HERE

Wednesday, October 15, 2014

EDIST - Edit distance

Edit distance

Given below code is for edist spoj or edit distance spoj .

Hint:-> Standerd edit distance problem you can read it from wikipidea.

1-> Using O(nm) space (n & m are size of strings )

#include <bits/stdc++.h>
using namespace std;
int dp[3000][3000];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char s[3000],t[3000];
        scanf("%s",s);
        scanf("%s",t);
        int size1=strlen(s),size2 = strlen(t);
        memset(dp,0,sizeof(dp[0][0])*(size1+1)*(size2+1));
        for(int i=0;i<=size1;i++)
            dp[i][0] = i;
        for(int j=0;j<=size2;j++)
            dp[0][j] = j;
        for(int i=1;i<=size1;i++)
        {
            for(int j = 1;j<=size2;j++)
            {
                if(s[i-1] == t[j-1]){
                    dp[i][j] = dp[i-1][j-1];
                }
                else
                    dp[i][j] = min(dp[i-1][j-1]+1,min(dp[i-1][j]+1,dp[i][j-1]+1));
            }
        }
        printf("%d\n",dp[size1][size2]);
    }
    return 0;
}


2-> Using O(n) space (n is size of strings )

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char s[3000],t[3000];
        scanf("%s",s);
        scanf("%s",t);
        int sizes = strlen(s),sizet = strlen(t);
        vector<int> v1(sizet+9),v2(sizet+9);
        for(int i=0;i<=sizet;i++)
            v1[i] = i;
        for(int i=0;i<sizes;i++)
        {
            v2[0] = i+1;
            for(int j=0;j<sizet;j++)
            {
                int add;
                if(s[i] == t[j])
                    add = 0;
                else 
                    add = 1;
                v2[j+1] = min(v2[j]+1,min(v1[j+1]+1,v1[j]+add));
            }
            for(int i=0;i<sizet;i++)
                v1[i] = v2[i];
        }
        printf("%d\n",v2[sizet]);
    }
    return 0;
}