pku 3110 贪心

pku 3110 贪心

贪心的思想首先要好好理解一下，这里给出了每一门考试开始的时间，和复习的最大提前时间（否则容易忘记），应该是枚举加贪心的思想

n的范围是50000，从大到小进行枚举，如果是考试的一天进堆，不是则找出最大的那个复习时间上限

比较难处理的是时间转化问题，我在这里是把它提前了31。1。1600作为一个起始时间，然后都做一个转换

第一个程序我是写了一个date的类，重载了各种操作，现在也觉得在这种比赛提里面如果没有必要还是少些类比较好，太复杂了，调试起来也不清晰

以下是ac代码：

#include < iostream >
#include < algorithm >
#define  maxn 50005
int  T[] = { 0 , 31 , 28 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31 };
using   namespace  std;
struct  node
{
     int  test,bg,i;
    friend  bool   operator < (node a,node b)
    {
         return  a.test > b.test;
    }
} in [maxn];
int  n;
bool  cmp2( int  x, int  y)
{
     return   in [x].bg < in [y].bg;
}
int  leap( int  y)
{
     if ((y % 4 == 0 && y % 100 ) || (y % 400 == 0 )) return   1 ;
     return   0 ;
}
int  DtoI( int  d, int  m, int  y)
{
     int  yy,mm,ret = 0 ;
     for (yy = 1600 ;yy < y;yy ++ )
        ret += 365 + leap(yy);
     for (mm = 1 ;mm < m;mm ++ )
        ret += T[mm] + ( int )(leap(y) && mm == 2 );
    ret += d;
     return  ret;
}
void  print( int  x)
{
     int  yy,mm,dd;
     for (yy = 1600 ;x >= 365 + leap(yy);yy ++ )
        {
            x -= 365 + leap(yy);
             if ( ! x)
                {
                    printf( " 31.12.%04d\n " ,yy);
                     return ;
            }
    }
     for (mm = 1 ;x >= T[mm] + ( int )(leap(yy) && mm == 2 );mm ++ )
        {
            x -= T[mm] + ( int )(leap(yy) && mm == 2 );
             if (x == 0 )
                {
                    printf( " %02d.%02d.%04d\n " ,T[mm] + ( int )(leap(yy) && mm == 2 ),mm,yy);
                     return ;
            }
    }
    printf( " %02d.%02d.%04d\n " ,x,mm,yy);
}
int  main()
{
     // freopen("1","r",stdin);
    scanf( " %d " , & n);
     int  i;
     for (i = 0 ;i < n;i ++ )
        {
             char  str[ 12 ];
            scanf( " %s " ,str);
             int  d,m,y,t;
            scanf( " %d.%d.%d " , & d, & m, & y);
            scanf( " %d " , & t);
             in [i].test = DtoI(d,m,y);
             in [i].bg = in [i].test - t;
    }
    sort( in , in + n);
     int  maxd = in [ 0 ].test,index =- 1 ;
     int  heap[maxn],heapsize = 0 ;
     bool  yes = true ;
     int  tmp = n;
     for (;yes && tmp;maxd -- )
        {
             if (maxd == in [index + 1 ].test)
                {
                    heap[heapsize ++ ] = index + 1 ;
                    push_heap(heap,heap + heapsize,cmp2);
                     ++ index;
                     continue ;
            }
             if (heapsize > 0 )
                {
                     int  top = heap[ 0 ];
                    pop_heap(heap,heap + heapsize,cmp2);
                    heapsize -- ;
                     if ( in [top].bg > maxd)yes = false ;
                     else  tmp -- ;
            }
     }
     if ( ! yes)
        printf( " Impossible\n " );
     else
        {
            print(maxd + 1 );
    }
     return   0 ;
 }