HDU - 1264 Counting Squares

最新推荐文章于 2019-12-28 13:25:20 发布

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最新推荐文章于 2019-12-28 13:25:20 发布

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分类专栏： ACM

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本文详细解析了一种通过矩形覆盖来计算单位正方形数量的算法。输入为一系列指定矩形角落坐标的行，输出为所有矩形覆盖的单位正方形总数。文章提供了一个C++实现示例，采用暴力法解决小规模数据集问题，关键在于避免重复计算被多个矩形覆盖的区域。

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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.

Input

The 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.

Output

Your 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

这道题本来应该是用线段树扫描线做的，但是看到数据量只有100*100，直接暴力就可以，细节还是很重要的

#include<bits/stdc++.h>
using namespace std;
bool vis[105][105];//判断一个点是否已经被覆盖了
int main()
{
    while(1)
    {
        memset(vis,false,sizeof(vis));
        int x1,x2,y1,y2;
        int flag=1;
        while(cin>>x1>>y1>>x2>>y2)
        {
            if(x1==-1&&x2==-1&&y1==-1&&y2==-1)
            {
                break;
            }
            if(x1==-2&&x2==-2&&y1==-2&&y2==-2)
            {
                flag=0;
                break;
            }
            for(int i=min(x1,x2);i<max(x1,x2);i++)//注意此地细节，因为对点横纵坐标大小关系不一样，所以要这样
            {
                for(int j=min(y1,y2);j<max(y1,y2);j++)
                {
                    vis[i][j]=true;
                }
            }
        }
        int sum=0;
        for(int i=0;i<=100;i++)
        {
            for(int j=0;j<=100;j++)
            {
                if(vis[i][j])
                {
                    sum++;//判断有多少个小的1*1的方格在里面，加起来就是面积
                }
            }
        }
        cout<<sum<<endl;
        if(flag==0)
        {
            break;
        }
    }
    return 0;
}