D

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[105][105];

int main()
{
    int a,b,c,d;
    memset(mp,0,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<0 || b<0 || c<0 || d<0)
        {
            int ans=0;
            for(int i=0;i<=100;++i)
            {
                for(int j=0;j<=100;++j)
                {
                    if(mp[i][j])
                    {
                        ++ans;
                    }
                }
            }
            printf("%d
",ans);
            memset(mp,0,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 0;
}