PROJECT EULER SOLUTION # 82

PROJECT EULER PROBLEM # 82

NOTE: This problem is a more challenging version of Problem 81.

The minimal path sum in the 5 by 5 matrix below, by starting in any cell in the left column and finishing in any cell in the right column, and only moving up, down, and right, is indicated in red and bold; the sum is equal to 994.

131 673 234 103 18 201 96 342 965 150 630 803 746 422 111 537 699 497 121 956 805 732 524 37 331

Find the minimal path sum, in matrix.txt (right click and 'Save Link/Target As...'), a 31K text file containing a 80 by 80 matrix, from the left column to the right column.

PROJECT EULER SOLUTION # 82

#include<stdio.h>

int mat[80][80],minimum[80][80],n;

int min2(int a,int b)

{

    return (a>b?b:a);

}

int min3(int a,int b,int c)

{

    return (a<b?(min2(a,c)):(min2(b,c)));

}

int main()

{

    int min,i,j;

    char temp;

    FILE *f;

    n=9;

    f=fopen("matrix.txt","r");

    if(f)

    {

        for(i=0;i<n;i++)

        {

            for(j=0;j<n-1;j++)

                fscanf(f," %d%c",&mat[i][j],&temp);

            fscanf(f,"%d",&mat[i][j]);

        }

        for(i=0;i<n;i++)

            minimum[i][0]=mat[i][0];

        for(i=1;i<n;i++)

        {

            j=0;

            minimum[j][i]=mat[j][i]+min2(minimum[j][i-1],minimum[j+1][i-1]+mat[j+1][i]);

            for(j=1;j<n;j++)

                minimum[j][i]=mat[j][i]+min2(minimum[j][i-1],minimum[j-1][i]);

            for(j=n-2;j>=0;j--)

                minimum[j][i]=min2(minimum[j+1][i]+mat[j][i],minimum[j][i]);

        }

        for(i=0;i<n;i++)

        {

            for(j=0;j<n;j++)

                printf("%d ",minimum[i][j]);

            printf("\n");

        }

        min=2147483647;

        for(i=0;i<n;i++)

            if(min>minimum[i][n-1])

                min=minimum[i][n-1];

        printf("answer = %d\n",min);

    }

    else

        printf("there is some error in opening the file\n");

    return 0;

}