UVA 836 - Largest Submatrix

UVA Sep 1, 2020

Problem PDF

Solution:

/******************************************************************
***   Problem       :                                           ***
***   Author        : Shipu Ahamed (Psycho Timekiller)          ***
***   E-mail        : [email protected]                   ***
***   University    : BUBT,Dept. of CSE                         ***
***   Team          : BUBT_Psycho                               ***
***   My Blog       : http://shipuahamed.blogspot.com           ***
***   Facebook      : http://www.facebook.com/DeesheharaShipu   ***
******************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define pi 2*acos(0.0)
#define all(v) v.begin(),v.end()

//input
#define si(t) scanf("%d",&t)
#define sl(t) scanf("%lld",&t)
#define sf(t) scanf("%f",&t)
#define sd(t) scanf("%lf",&t)
#define sc(c) scanf("%c",&c)
#define sii(a,b) scanf("%d%d",&a,&b)
#define sll(a,b) scanf("%lld%lld",&a,&b)

//Output
#define P(a) printf("%d\n",a)
#define PL(a) printf("%lld\n",a)
#define PF(a) printf("%f\n",a)
#define PD(a) printf("%lf\n",a)
#define PS(a) printf("%s\n",a)
#define PSN(a) printf("%s ",a)
#define PN(a) printf("%d ",a)
#define PLN(a) printf("%lld ",a)
#define PFN(a) printf("%f ",a)
#define PDN(a) printf("%lf ",a)
#define PP(a,b) printf("%d %d\n",a,b)
#define PPN(a,b) printf("%d %d ",a,b)
#define PPL(a,b) printf("%lld %lld\n",a,b)
#define PPLN(a,b) printf("%lld %lld ",a,b)

#define CP(a) cout< vi;
typedef  vector vll;
typedef  vector vs;
typedef  set si;
typedef  set ss;
typedef  map mii;
typedef  map mll;
typedef  map msi;
typedef  map mci;

template string toString( T Number ){stringstream st;st << Number;return st.str();}
template  T SOD(T n) {__typeof(n) sum=0;for(__typeof(n) i=1;i*i<=n;i++)sum+=(n%i)?0:((i*i==n)?i:i+n/i);return sum;}

//For Define
#define forab(i,a,b) for(__typeof(b) i=(a);i<=(b);i++)
#define for0(i,n) forab(i,0,(n)-1)
#define for1(i,n) forab(i,1,n)
#define rforab(i,b,a) for(__typeof(b) i=(b);i>=(a);i--)
#define rfor0(i,n) rforba(i,(n)-1,0)
#define rfor1(i,n) rforba(i,n,1)
#define forstl(i,s) for(__typeof((s).end()) i=(s).begin(); i != (s).end(); i++)

//Debug
#define dbg(x) cout << #x << " -> " << (x) << endl;
#define dbgsarr(i,a) cout<<#a<<"["< "< "< "<>p ; return p;}

ll pow(ll a,ll b, ll m) { ll res = 1; while(b) { if(b & 1) { res = ( (res % m) * (a % m) ) %m ; } a= ((a%m) * (a%m)) %m; b >>= 1; } return res; }
ll modInverse(ll a, ll m){return pow(a,m-2,m);}

////============ CONSTANT ===============////
#define mx7   10000007
#define mx6   1000006
#define mx5   100005
#define inf   1<<30                                           //infinity value
#define eps   1e-9
#define mx    (100010)
#define mod   1000000007
////=====================================////

struct zeone
{
    int one=0,zero=0;

};

int main()
{
    int t;
    si(t);
    while(t--)
    {
        string s[30];
        cin>>s[0];

        int n=sz(s[0]);

        zeone cs[50][50];

        for0(i,n)
        {
            if(s[0][i]=='0')
            {
                cs[1][i+1].zero = cs[1][i].zero+1;
                cs[1][i+1].one = cs[1][i].one;
            }
            else
            {
                cs[1][i+1].one = cs[1][i].one+1;
                cs[1][i+1].zero = cs[1][i].zero;
            }

//            PN(cs[1][i+1].one);
        }
//        nl;

        for1(i,n-1)
        {
            cin>>s[i];
            for0(j,n)
            {
                if(s[i][j]=='0')
                {
                    cs[i+1][j+1].zero = cs[i][j+1].zero + cs[i+1][j].zero - cs[i][j].zero +1;
                    cs[i+1][j+1].one = cs[i][j+1].one + cs[i+1][j].one - cs[i][j].one;
                }
                else
                {
                    cs[i+1][j+1].zero = cs[i][j+1].zero + cs[i+1][j].zero - cs[i][j].zero;
                    cs[i+1][j+1].one = cs[i][j+1].one + cs[i+1][j].one - cs[i][j].one+1;
                }
//                PN(cs[i+1][j+1].one);

            }
//            nl;
        }

//        nl;
//        for0(i,n)
//        {
//            cout<=p)
                        {
                            res.zero = p;
                            p = ( cs[i][j].one - (cs[i][l-1].one + cs[k-1][j].one) ) + cs[k-1][l-1].one;
                            if(res.one
https://github.com/Shipu/OnlineJudgeProblemSolutionWithCPlusPlus/tree/master/uva/836/836.cpp

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