D - Counting Squares

Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line 
5 8 7 10 
specifies the rectangle who's corners are(5,8),(7,8),(7,10),(5,10). 
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once. 

InputThe input format is a series of lines, each containing 4 integers. Four -1's are used to separate problems, and four -2's are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of a rectangle. 
OutputYour output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line. 
Sample Input

5 8 7 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2

Sample Output

8
10000

数据太小了，直接暴力

 #include <cstdio>
 #include <cstring>
 #include <iostream>

 using namespace std;

 bool mp[][];

 int main()
 {
     int a,b,c,d;
     memset(mp,,sizeof(mp));
     while(scanf("%d %d %d %d",&a,&b,&c,&d)!=EOF)
     {
         // 每一组结束，输出并更新
         if(a< || b< || c< || d<)
         {
             int ans=;
             for(int i=;i<=;++i)
             {
                 for(int j=;j<=;++j)
                 {
                     if(mp[i][j])
                     {
                         ++ans;
                     }
                 }
             }
             printf("%d\n",ans);
             memset(mp,,sizeof(mp));
             continue;
         }
         // 暴力覆盖
         if(a>c)
         {
             swap(a,c);
         }
         if(b>d)
         {
             swap(b,d);
         }
         for(int i=a;i<c;++i)
         {
             for(int j=b;j<d;++j)
             {
                 mp[i][j]=true;
             }
         }
     }

     return ;
 }