We are given a figure consisting of only horizontal and vertical line segments. Our goal is to count the number of all different rectangles formed by these segments. As an example, the number of rectangles in the Figures 1 and 2 are 5 and 0 respectively.

There are many intersection points in the figure. An intersection point is a point shared by at least two segments. The input line segments are such that each intersection point comes from the intersection of exactly one horizontal segment and one vertical segment.

The first line of the file contains a single number M, which is the number of test cases in the file (1 <= M <= 10), and the rest of the file consists of the data of the test cases. Each test case begins with a line containing s (1 <= s <= 100), the number of line segments in the figure. It follows by s lines, each containing x and y coordinates of two end points of a segment respectively. The coordinates are integers in the range of 0 to 1000.

The above input file contains two test cases corresponding to Figures 1 and 2 respectively.

题目大意：给一些水平或竖直的线段，求能组成的矩形的个数。

解题思路：因为题目给的只有垂直和水平的线段，且总线段不超过100.所以我们可以暴力。

　　1、任选两根水平的线段，若无水平线段可选，结束。否则，转2

　　2、从所有的垂直线段里，找到和这两根水平线段相交的线段，假设有tmp条。转3

　　3、对于1步选的两条水平线段，因为有tmp跟垂直线段与其相交，根据推算，可以得知，其能组成的矩形就是（tmp - 1）*tmp / 2 个，将其加进总和里即可。转1

 1 #include<iostream>
 2 #include<string.h>
 3 using namespace std;
 4 class Rect{
 5 public:
 6     int x1,y1,x2,y2;
 7     void set(int a,int b,int c,int d){
 8         x1=a,y1=b,x2=c,y2=d;
 9     }
10 };//线段类
11 bool ok(Rect &a,Rect &b){
12     return b.y1<=a.y1 && a.y1<=b.y2 && a.x1<=b.x1 && b.x1<=a.x2;
13 }//判断线段相交
14 int M;
15 int s;
16 Rect rectH[1004],rectS[1004];//水平和竖直线段集
17 int main(){
18     cin>>M;
19     while(M--){
20         cin>>s;
21         int H=0,S=0;
22         for(int i=0;i<s;i++){
23             int x,y,x1,y1;
24             cin>>x>>y>>x1>>y1;
25             if(x==x1){
26                 if(y>y1)rectS[S++].set(x1,y1,x,y);
27                 else rectS[S++].set(x,y,x1,y1);
28             }else{
29                 if(x>x1)rectH[H++].set(x1,y1,x,y);
30                 else rectH[H++].set(x,y,x1,y1);
31             }//要注意从上到下，从左到右
32         }
33 
34         int tot=0;
35         for(int i=0;i<H-1;i++){
36             for(int j=i+1;j<H;j++){//枚举2条横的，统计满足相交的竖着的线段的条数count
37                 int count=0;
38                 for(int k=0;k<S;k++){
39                     if(ok(rectH[i],rectS[k]) && ok(rectH[j],rectS[k]))
40                         count++;
41                 }
42                 tot+=(count-1)*count/2;//计算此情况能组成多少
43             }
44         }
45         cout<<tot<<'\n';
46     }return 0;
47 }

本文转自beautifulzzzz博客园博客，原文链接：http://www.cnblogs.com/zjutlitao/p/3572395.html，如需转载请自行联系原作者