2018暑期SICNU-ACM组集训报告（1）

题目：

Xzz need to calculate Intersection over Union(IoU) of two rectangles, can you help him?

rectangle (x, y, w, h) means a rectangle MNPQ, M(x,y), N(x, y+h), P(x+w, y+h), Q(x+w, y).

IoU = Area of overlap / Area of union.

Input

First line of the input file contains an integer T(0 < T <= 100) that indicates how many cases of inputs are there.

The description of each case is given below:

The first line of each input set contains integer x1, y1, w1, h1.

The second line of each input set contains integer x2, y2, w2, h2.

0 ≤ x, y, w, h ≤ 100000

Output

The description of output for each test case is given below:

The first line of the output for each test case contains number k- the IoU of two rectangles.

Output should be rounded to 2 digits after decimal point.

Sample Input

2

1 1 1 1

1 1 2 2

1 1 2 1

1 1 1 2

Sample Output

0.25

0.33

原题链接

题意：
输入两组各四个数据按照一定的方式计算为两个矩形的四点坐标,计算两矩形重合面积和合并面积的比值
解题思路：
要算两个面积就分开讨论：
重合面积：
由于是矩形，所以知道两个对角点的坐标就可以计算重合面积的大小了。
首先在横纵坐标中找到两个中间值也就是一共四个值，用它来分别表示重合面积四个边在坐标系中的横纵坐标的值，再相减得到重合矩形的长宽，即可得到该面积。

合并面积：直接运算，相加后再减去重合面积即可

这道题还是有坑点，要考虑是否相交。写完了也没检查直接交了一发，WA后检查发现在计算合并面积之后没有减去多加的重合面积（小坑点，但是自己不检查这个习惯真的很大，太自信了）

AC代码：

#include<iostream>
#include<cstdio>
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
double area_o;
using namespace std;
double area_over(int m1,int n1,int m2,int n2)
{
    
    if(m2>m1 && n2>n1)
        area_o=(m2 -m1)*(n2-n1);
    else
        area_o=0.00;
    return 0;
}


int main(int argc, char const *argv[])
{
    int n;
    cin>>n;
    double m1,n1,m2,n2;
    double area1=0,area2=0,area_union=0;
    double IoU=0;
    int x1,x2,y1,y2,w1,w2,h1,h2;
    for(int i=0;i<n;i++){
        cin>>x1>>y1>>w1>>h1;
        cin>>x2>>y2>>w2>>h2;
        m1 = max(min(x1,x1+w1),min(x2,x2+w2)); 
        n1 = max(min(y1+h1,y1),min(y2,y2+h2));  
        m2 = min(max(x1,x1+w1),max(x2,x2+w2));
        n2 = min(max(y1+h1,y1),max(y2,y2+h2));
        area_over(m1,n1,m2,n2);
        area1=w1*h1;area2=w2*h2;
        area_union=area1+area2;
        IoU=(area_o*1.0)/((area_union-area_o)*1.0);
        printf("%.2lf\n",IoU);
    }
    return 0;
}