hdu_3512

题意：最小二分图匹配的变形，边的权值为两定点的乘积，求匹配中的最大乘积最小。

分析：首先，两组数据中全为正时，只有大乘小才能得到最大中的最小。
明白这点后，分析两组数据的情况，都为正，都为负，一负一正，部分负部分正，考虑清楚，并想出对应的解法即可。还有就是0既可以看成负 数，也可看成正数。

#include <cstdio>
#include  <stdlib.h>
#include  <algorithm>
using namespace  std;
const long long  inf = -1LL<<60 ;

long long a[100010],b[100010 ];
int  xa,xb,ya,yb;
int  n;

int  main()
{  
    int  i,j,k;
    long long  max;
    while (scanf("%d",&n) !=  EOF)
    {
          max =  inf;
          
          xa = xb = ya = yb = 0 ;
          for (i = 1; i <= n; i++ )
          {
             scanf("%I64d",& a[i]);
             if (a[i] <= 0 )
                ya ++ ;
             else xa ++ ;
          }
          for (i = 1; i <= n; i++ )
          {
              scanf("%I64d",& b[i]);
              if (b[i] <= 0 )
                 yb ++ ;
              else xb ++ ;
          }
          sort(a+1,a+1+ n);
          sort(b+1,b+1+ n);
          //a，b全为正 
          if ((xa == n && xb == n) ||(xa == 0 && xb == 0 ))
          {
                  for (i = 1; i <= n; i++ )
                      if (max < a[i]*b[n-i+1 ])
                         max = a[i]*b[n-i+1 ];
                  printf("%I64d\n" ,max);
                  continue ;
          }
          //a，b一正一负 
          if ((xa == n && xb == 0) || (xa == 0 && xb ==  n))
          {
                  for (i = 1; i <= n; i++ )
                      if (max < a[i]* b[i])
                         max = a[i]* b[i];
                  printf("%I64d\n" ,max);
                  continue ;
          }
          //a全为正
          if (xa ==  n)
          {
                 for (i = 1; i <= xb;i++ )
                     if (max < a[i]*b[n-i+1 ])
                       max = a[i]*b[n-i+1 ];
                 printf("%I64d\n" ,max);
                 continue ;
          } 
          //a全为负 
          if (xa == 0 )
          {
                 for (i = 1; i <= yb; i++ )
                     if (max < a[n-i+1]* b[i])
                        max = a[n-i+1]* b[i];
                printf("%I64d\n" ,max);
                continue ;
          }
           //b全为正
          if (xb ==  n)
          {
                 for (i = 1; i <= xa;i++ )
                     if (max < b[i]*a[n-i+1 ])
                       max = b[i]*a[n-i+1 ];
                 printf("%I64d\n" ,max);
                 continue ;
          } 
          //b全为负 
          if (xb == 0 )
          {
                 for (i = 1; i <= ya; i++ )
                     if (max < b[n-i+1]* a[i])
                        max = b[n-i+1]* a[i];
                printf("%I64d\n" ,max);
                continue ;
          }
          //a的正==b的负 
          if (xa ==  yb)
          {
                 for (i = 1; i <= ya; i++ )
                     if (max < a[i]*b[yb+ i])
                        max = a[i]*b[yb+ i];
                 for (i = 1; i <= yb; i++ )
                     if(max < b[i]*a[ya+ i])
                            max = b[i]*b[yb+ i];       
                 printf("%I64d\n" ,max);
                 continue ;
          }
          //a的正 》b的负，余下全为正 
          if (xa >  yb)
          {
                 for (i = ya+1,j = n - ya; i <= n-yb; i++,j-- )
                     if (max <a[i]* b[j])
                        max = a[i]* b[j];
                 printf("%I64d\n" ,max);
                 continue ; 
          } 
          //a的正小于b的负，余下全为负 
          if (xa <  yb)
          {
                 for (i = xb+1,j = n- xb; i <= n - xa; i++,j-- )
                     if (max < a[i]* b[j])
                        max = a[i]* b[j];
                 printf("%I64d\n" ,max);
                 continue ;
          }           
    }
    return 0 ;
}