这道题是离散化的入门题。
问题描述：一些已知右下顶点和左上顶点坐标的矩形，这些矩形可能部分重叠，求它们所占的实际面积。
方案一、
把矩形映射到数组M[ ][ ]中，如果某矩形的位置为（x1，y1）（x2，y2）则M[i][j]=1（其中x1<=i<=x2,y1<=j<=y2）。但是，如果坐标值离散程度很大或者非整型，此方案就不可行。
方案二、
判断任两个矩形是否相交，如果重叠，两面积相加，减去重叠部分。但重叠情况很多，算法不复杂但程序写起来很复杂，容易出错。
方案三——离散化
1、首先分离出所有的横坐标和纵坐标分别按升序存入数组X[ ]和Y[ ]中.
2、 设数组XY[ ][ ].对于每个矩形(x1,y1)(x2,y2)确定i1,i2,j1,j2,使得,X[i1]>x1,X[i2]<=x2,Y[i1]>y1,Y[i2]>=y2令XY[ i ][ j ] = 1 (i从i1到i2，j从j1到j2)
3、统计面积：area+=XY[i][j] *(X[i]-X[i-1])*(Y[i] – Y[i-1])

#include < iostream >
using   namespace  std;
 double  x[ 201 ],y[ 201 ],s[ 101 ][ 4 ];
 int  xy[ 201 ][ 201 ];
 int  n,cas = 0 ;
 double  sum;
 int  main()
{
     int  i,j,k;
     while (cin >> n)
    {   
         if (n == 0 )
             break ;
        cas ++ ;
        k = 0 ;
        sum = 0.0 ;
        memset(xy, 0 , sizeof (xy));
         for (i = 1 ;i <= n;i ++ )
        {
            cin >> s[i][ 0 ] >> s[i][ 1 ] >> s[i][ 2 ] >> s[i][ 3 ];
            x[k] = s[i][ 0 ];
            y[k] = s[i][ 1 ];
            k ++ ;
            x[k] = s[i][ 2 ];
            y[k] = s[i][ 3 ];
            k ++ ;
        }
        sort(x,x + 2 * n);
        sort(y,y + 2 * n);
         for (k = 1 ;k <= n;k ++ )
        {
             int  i1,i2,j1,j2;
             for (i1 = 0 ;i1 < 2 * n;i1 ++ )
            {
                 if (x[i1] == s[k][ 0 ])
                     break ;
            }
             for (i2 = 0 ;i2 < 2 * n;i2 ++ )
            {
                 if (x[i2] == s[k][ 2 ])
                     break ;
            }
             for (j1 = 0 ;j1 < 2 * n;j1 ++ )
            {
                 if (y[j1] == s[k][ 1 ])
                     break ;
            }
             for (j2 = 0 ;j2 < 2 * n;j2 ++ )
            {
                 if (y[j2] == s[k][ 3 ])
                     break ;
            }
             for (i = i1;i < i2;i ++ )
            {
                 for (j = j1;j < j2;j ++ )
                {
                    xy[i][j] = 1 ;
                }
            }
        }
         for (i = 0 ;i < 2 * n;i ++ )
        {
             for (j = 0 ;j < 2 * n;j ++ )
            {
                sum += xy[i][j] * (x[i + 1 ] - x[i]) * (y[j + 1 ] - y[j]);
            }
        }
        printf( " Test case #%d\n " ,cas);
        printf( " Total explored area: %.2f\n " ,sum);
        printf( " \n " );
    }
    system( " pause " );
     return   0 ;
}