Cod sursa(job #2312627)

Utilizator usureluflorianUsurelu Florian-Robert usureluflorian Data 5 ianuarie 2019 11:28:28
Problema DreptPal Scor 40
Compilator cpp-64 Status done
Runda Arhiva de probleme Marime 2.08 kb
#include <bits/stdc++.h>
using namespace std;
class InParser {
private:
	FILE *fin;
	char *buff;
	int sp;

	char read_ch() {
		++sp;
		if (sp == 4096) {
			sp = 0;
			fread(buff, 1, 4096, fin);
		}
		return buff[sp];
	}

public:
	InParser(const char* nume) {
		fin = fopen(nume, "r");
		buff = new char[4096]();
		sp = 4095;
	}

	InParser& operator >> (int &n) {
		char c;
		while (!isdigit(c = read_ch()) && c != '-');
		int sgn = 1;
		if (c == '-') {
			n = 0;
			sgn = -1;
		} else {
			n = c - '0';
		}
		while (isdigit(c = read_ch())) {
			n = 10 * n + c - '0';
		}
		n *= sgn;
		return *this;
	}

	InParser& operator >> (long long &n) {
		char c;
		n = 0;
		while (!isdigit(c = read_ch()) && c != '-');
		long long sgn = 1;
		if (c == '-') {
			n = 0;
			sgn = -1;
		} else {
			n = c - '0';
		}
		while (isdigit(c = read_ch())) {
			n = 10 * n + c - '0';
		}
		n *= sgn;
		return *this;
	}
};
ofstream g ("dreptpal.out");
const int nmax=1e3+3;
int n,m,v[nmax][nmax],d[nmax][nmax],c,r,mr,sol;
stack <int> st;
int main()
{
    InParser f ("dreptpal.in");
    f>>n>>m;
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=m;++j)
        {
            f>>v[i][j];
            ++v[i][j];
        }
    }
    for(int i=1;i<=n;++i)
    {
        c=r=0;
        for(int j=1;j<=m;++j)
        {
            if(r>=j) mr=2*c-j;
            d[i][j]=min(m-j,d[i][mr]);
            d[i][j]=max(d[i][j],1);
            while(v[i][j-d[i][j]]==v[i][j+d[i][j]]&&j-d[i][j]>=1&&j+d[i][j]<=m) ++d[i][j];
            if(i+d[i][j]-1>r)
            {
                c=i;
                r=i+d[i][j]-1;
            }
        }
    }
    for(int j=1;j<=m;++j)
    {
        sol=max(sol,d[1][j]*2-1);
        st.push(1);
        for(int i=2;i<=n+1;++i)
        {
            while(!st.empty()&&(d[i][j]<d[st.top()][j]))
            {
                sol=max(sol,(i-st.top())*(d[st.top()][j]*2-1));
                st.pop();
            }
            st.push(i);
        }
        while(!st.empty()) st.pop();
    }
    g<<sol;
    return 0;
}