Light OJ: 1424 – New Land

Problem Link :

Solution Idea: I think this problem can be solved with a 2D Segment Tree. But here I use the histogram technique to solve this problem. You can try Spoj – HISTOGRA – Largest Rectangle in a Histogram and Light OJ – 1083 – Histogram . Both are same problem on different website. This problems can be using a niche technique called histogram which can be implemented using a stack. the algorithm determine the index of the bar before the current bar which is smaller than the current bar also determine the index of the bar after the current bar which is smaller than the current bar. Here the function func is the answer of the above two problem.

In this problem for each row we consider a histogram from current row and the rows which are above the current row with continuous 0. Then determine the maximum answer.

#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}

#define mx 2002

char grid[mx][mx];

int dp[mx][mx];

int func(int ara[], int n)
    int left[mx];


    for(int i=0; i<n; i++)
        while(!st.empty() && ara[]>=ara[i]) st.pop();

    while(!st.empty()) st.pop();

    int ret=0;
    int right;
    for(int i=n-1; i>=0; i--)
        while(!st.empty() && ara[]>=ara[i]) st.pop();
    return ret;

int main()

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    int t;
        int n,m;
        for(int i=0; i<n; i++)

        for(int j=0; j<m; j++)

        int ans=func(dp[0],m);

        for(int i=1; i<n; i++)
            for(int j=0; j<m; j++)


    return 0;