UVa 10074 - Take the Land Solution

#include<iostream>
#include<cstdio>
#include<list>
#include<string>
#include<cstring>
#include<sstream>
#include<cctype>
#include<string.h>
#include<algorithm>
#include<cmath>
#include<stack>
#include<fstream>
#include<cstdlib>
#include<vector>
#include<map>
#include<set>
#include<utility>
#include<iomanip>
#include<queue>
#include<deque>
#include<iterator>
#include<assert.h>
#include<bitset>
#include<climits>
#include<ctime>
#include<complex>

using namespace std;

#define SET(a) memset(a,-1,sizeof(a))
#define ALL(a) a.begin(),a.end()
#define CLR(a) memset(a,0,sizeof(a))
#define PB push_back
#define PI acos(-1.0)
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define READ freopen("input.txt", "r", stdin)
#define WRITE freopen("output.txt", "w", stdout)
#define LL long long

#define S(a) scanf("%d",&a)
#define S2(a,b) scanf("%d%d",&a,&b)
#define KS printf("Case %d: ",kk++)

#define MOD 1000000007
#define MX 100010


int row[110][110],mat[110][110],n,m;

int func(int start, int end)
{
    int i,j,curnt=0, mx=-1270000;
    for(i=1;i<=n;i++)
    {
        j=row[i][end]-row[i][start-1];
        curnt+=j;
        if(curnt>mx) mx=curnt;
        if(curnt<0) curnt=0;
    }
return mx;
}

int main()
{
    int i,j,temp;
    while(cin>>n>>m)
    {
        if(!n && !m) break;
        for(i=0;i<=n;i++)
            row[i][0]=0;


        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                cin>>mat[i][j];
                if(mat[i][j]==0)
                    mat[i][j]=1;
                else mat[i][j]=-1e5;
                row[i][j]=row[i][j-1]+mat[i][j];
            }

        temp=0;
        for(i=1;i<=m;i++)
            for(j=i;j<=m;j++)
                temp=max(temp,func(i,j));

        cout<<temp<<endl;
        //if(tc) cout<<endl;
    }
return 0;
}