Seat taking up is tough (hdu 3262 easy simulation questions

http://acm.hdu.edu.cn/showproblem.php?pid=3262

Meaning of the questions: accounting location must continuously find the most comfortable value left in this condition maximum (because of their own to sit)

If you account for less than p continuous then there is only an own position

Account for less than output -1

Solution: it is simple for loop line by line, looking past a row there (the number of columns - the number of successive positions +1 to account for the) kind of continuous, each investigation in the past if we are all empty position can be accounted for, but also to set up maxx found maximum

#include<bits/stdc++.h>
using namespace std;
struct node
{
    int idd,time,h,m,q;
}a[55];
bool cmp(node x,node y)
{
    return x.time<y.time;
}
int mp[55][55];
int main()
{
    int i,j,n,m,k;
    while(scanf("%d%d%d",&n,&m,&k),n||m||k)
    {
        memset(mp,-1,sizeof(mp));
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%d",&mp[i][j]);
            }
        }
        for(i=1;i<=k;i++)
        {
            scanf("%d:%d %d",&a[i].h,&a[i].m,&a[i].q);
            a[i].idd=i;
            a[i].time=a[i].h*60+a[i].m;
        }
        sort(a+1, A + . 1 + K, CMP); // Note After sorting output to the original input sequentially outputs 
        Vector < int > VE [ 55 ];
         for (I = . 1 ; I <= K; I ++ ) 
        { 
            int P = A [I ] .q;
             int Maxx = - 0x3f3f3f3f ;
             int maxydw1 = 0 , maxydw2 = 0 ;
             for (J = . 1 ; J <= n-; J ++ ) 
            { 
                int GS = m-P + . 1 ;
                 for ( int O = . 1 ; O < = gs; o ++) 
                { 
                    Int F = . 1 ;
                     for ( int Ql = O; Ql <= p-O + . 1 ; Ql ++) // start wrong here for a long time 
                    {
                         IF (MP [J] [Ql] == - . 1 ) F = 0 ; 
                    } 
                    IF (F == . 1 && MP [J] [O]> Maxx) 
                    { 
                        Maxx = MP [J] [O]; 
                        maxydw1 = J; 
                        maxydw2 = O; 
                    } 
                }
            }
            if(maxydw1!=0)
            {
                int po=maxydw2+p-1;
                for(int o=maxydw2;o<=po;o++)
                {
                    mp[maxydw1][o]=-1;
                }
                ve[a[i].idd].push_back(maxydw1);
                ve[a[i].idd].push_back(maxydw2);
            }
            else
            {
                int mx=0,my=0,maxxx=-0x3f3f3f3f;
                for(int q1=1;q1<=n;q1++)
                {
                    for(int q2=1;q2<=m;q2++)
                    {
                        if(mp[q1][q2]!=-1&&mp[q1][q2]>maxxx)
                        {
                            maxxx=mp[q1][q2];
                            mx=q1,my=q2;
                        }
                    }
                }
                if(mx!=0)ve[a[i].idd].push_back(mx),ve[a[i].idd].push_back(my),mp[mx][my]=-1;
                else ve[a[i].idd].push_back(-1);
            }
        }
        for(i=1;i<=k;i++)
        {
            if(ve[i].size()==1)cout<<-1<<endl;
            else cout<<ve[i][0]<<" "<<ve[i][1]<<endl;
        }
    }
    return 0;
}

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