来个非算法的，求矩形交集的面积

[healer_kx]

在一个平面坐标系上，有两个矩形，它们的边分别平行于X和Y轴。
其中，矩形A已知， ax1(左边), ax2（右边）, ay1（top的纵坐标）, ay2（bottom纵坐标）. 矩形B，类似，就是 bx1, bx2, by1, by2。这些值都是整数就OK了。
要求是，如果矩形没有交集，返回-1， 有交集，返回交集的面积。

int area(rect const& a, rect const& b)
{
...
}

补齐代码，我认为好的代码应该是简洁的。别用库。你可以写你的辅助函数，宏定义。代码风格也很重要。

[xiao910xx]

选两个矩形的左上角的坐标 可判断出上下和左右的位置关系
然后可选择 一个的左上的坐标 一个选择右下的坐标 可得到是否重叠或重叠矩形的边长

[赵4老师]

68楼中指令
004010A7 FF 24 9D 40 12 40 00 jmp dword ptr [ebx*4+401240h]
使用的跳转表：

00401240  00401234  004010AE  004010CA
0040124C  004010D1  004010ED  00401226
00401258  004010F4  00401110  0040112F
00401264  0040114B  0040116A  00401180
00401270  00401199  004011AF  004011C8
0040127C  004011DE  004011F7  0040120D

[赵4老师]

编译运行通过的版本：

#include <stdio.h>
int ax1,ay1,ax2,ay2;
int bx1,by1,bx2,by2;
//在一个平面坐标系上，有两个矩形，它们的边分别平行于X和Y轴。
//其中，矩形A已知， ax1(左边), ax2（右边）, ay1（top的纵坐标）, ay2（bottom纵坐标）. 矩形B，类似，就是 bx1, bx2, by1, by2。这些值都是整数就OK了。
//要求是，如果矩形没有交集，返回-1， 有交集，返回交集的面积。
int area() {
    int b1c=0,b2c=0;

    //__asm int 3;//快速定位Release版汇编代码用
         if (bx1<ax1)  b1c+=3*0;
    else if (ax2<=bx1) b1c+=3*2;
    else               b1c+=3*1;

         if (by1<ay1)  b1c+= 0;
    else if (ay2<=by1) b1c+= 2;
    else               b1c+= 1;

         if (bx2<ax1)  b2c+=3*0;
    else if (ax2<=bx2) b2c+=3*2;
    else               b2c+=3*1;

         if (by2<ay1)  b2c+= 0;
    else if (ay2<=by2) b2c+= 2;
    else               b2c+= 1;

    switch (b1c*9+b2c) {
    case 0:
    case 1:
    case 2:
    case 3:
    case 6:
    case 10:
    case 11:
    case 20:
    case 23:
    case 26:
    case 30:
    case 33:
    case 50:
    case 53:
    case 60:
    case 61:
    case 62:
    case 70:
    case 71:
    case 80:
        return -1;
    case 4:
        return (bx2-ax1)*(by2-ay1);
    case 5:
        return (bx2-ax1)*(ay2-ay1);
    case 7:
        return (ax2-ax1)*(by2-ay1);
    case 8:
        return (ax2-ax1)*(ay2-ay1);
    case 13:
        return (bx2-ax1)*(by2-by1);
    case 14:
        return (bx2-ax1)*(ay2-by1);
    case 16:
        return (ax2-ax1)*(by2-by1);
    case 17:
        return (ax2-ax1)*(ay2-by1);
    case 31:
        return (bx2-bx1)*(by2-ay1);
    case 32:
        return (bx2-bx1)*(ay2-ay1);
    case 34:
        return (ax2-bx1)*(by2-ay1);
    case 35:
        return (ax2-bx1)*(ay2-ay1);
    case 40:
        return (bx2-bx1)*(by2-by1);
    case 41:
        return (bx2-bx1)*(ay2-by1);
    case 43:
        return (ax2-bx1)*(by2-by1);
    case 44:
        return (ax2-bx1)*(ay2-by1);
    case 9:
    case 12:
    case 15:
    case 18:
    case 19:
    case 21:
    case 22:
    case 24:
    case 25:
    case 27:
    case 28:
    case 29:
    case 36:
    case 37:
    case 38:
    case 39:
    case 42:
    case 45:
    case 46:
    case 47:
    case 48:
    case 49:
    case 51:
    case 52:
    case 54:
    case 55:
    case 56:
    case 57:
    case 58:
    case 59:
    case 63:
    case 64:
    case 65:
    case 66:
    case 67:
    case 68:
    case 69:
    case 72:
    case 73:
    case 74:
    case 75:
    case 76:
    case 77:
    case 78:
    case 79:
        printf("not ax1<=ax2 or not ay1<ay2 or not bx1<bx2 or not by1<by2!");
        return -1;
    break;
    }
    return -1;
}
int main() {
    int a;

    ax1=20;ay1=30;
    ax2=80;ay2=70;
    bx1=10;by1=50;
    bx2=90;by2=60;
    a=area();
    printf("area=%d\n",a);
}
//area=600

[ruf]

bool intersect(const SkIRect& a, const SkIRect& b) {
SkASSERT(&a && &b);

if (!a.isEmpty() && !b.isEmpty() &&
a.fLeft < b.fRight && b.fLeft < a.fRight &&
a.fTop < b.fBottom && b.fTop < a.fBottom) {
fLeft = SkMax32(a.fLeft, b.fLeft);
fTop = SkMax32(a.fTop, b.fTop);
fRight = SkMin32(a.fRight, b.fRight);
fBottom = SkMin32(a.fBottom, b.fBottom);
return true;
}
return false;
}

[ri_aje]

#include <cstdio>

template <typename T> T const& min (T const& x, T const& y) { return x<y ? x : y; }
template <typename T> T const& max (T const& x, T const& y) { return x>y ? x : y; }

int ax0,ay0,ax1,ay1;
int bx0,by0,bx1,by1;

int area ()
{
 int const dx = min(ax1,bx1) - max(ax0,bx0);
 int const dy = min(ay1,by1) - max(ay0,by0);
 return dx>=0&&dy>=0 ? dx*dy : -1;
}

int main ()
{
 int a;

 ax0=20;ay0=30;
 ax1=80;ay1=70;
 bx0=10;by0=50;
 bx1=90;by1=60;
 a=area();
 printf("area=%d\n",a);

 return 0;
}

	movl	ax1, %eax
	movl	ax0, %edx
	cmpl	%eax, bx1
	cmovle	bx1, %eax
	cmpl	%edx, bx0
	cmovge	bx0, %edx
	movl	ay1, %ecx
	subl	%edx, %eax
	movl	ay0, %edx
	cmpl	%ecx, by1
	cmovle	by1, %ecx
	cmpl	%edx, by0
	cmovge	by0, %edx
	subl	%edx, %ecx
	js	.L3
	testl	%eax, %eax
	js	.L3
	imull	%ecx, %eax
	ret
.L3:
	movl	$-1, %eax
	ret

[yuriarthas]

怎么楼上这么多代码？不就是看两个矩阵是否有重合的地方么？怎么这么麻烦？还是我想错了？

[老王爱上猫]

当初面试的时候没有弄出来

[niu_a]

int longth(int first, int second)
{
return second-first;
}
int min(int first, int second)
{
return first<second?first:second;
}
int max(int first,int second)
{
return first<second?second:first;
}
int area(rect const &a, rect const&b)
{
int w = longth(a.left,a.right)+longth(b.left,b.right)-longth(min(a.left,b.left),max(a.right,b.right));
if(w>0)
{
int h=longth(a.top,a.bottom)+longth(b.top,b.bottom)-longth(min(a.top,b.top),max(a.bottom,b.bottom));
if(h>0)
return w*h；
}
return 0;
}

[niu_a]

int longth(int first, int second)
{
return second-first;
}
int min(int first, int second)
{
return first<second?first:second;
}
int max(int first,int second)
{
return first<second?second:first;
}
int area(rect const &a, rect const&b)
{
int w = longth(a.left,a.right)+longth(b.left,b.right)-longth(min(a.left,b.left),max(a.right,b.right));
if(w>0)
{
int h=longth(a.top,a.bottom)+longth(b.top,b.bottom)-longth(min(a.top,b.top),max(a.bottom,b.bottom));
if(h>0)
return w*h；
}
return 0;
}

[niu_a]

int longth(int first, int second)
{
	return second-first;
}
int min(int first, int second)
{
	return first<second?first:second;
}
int max(int first,int second)
{
	return first<second?second:first;
}
int area(rect const &a, rect const&b)
{
	int w = longth(a.left,a.right)+longth(b.left,b.right)-longth(min(a.left,b.left),max(a.right,b.right));
	if(w>0)
	{
		int h=longth(a.top,a.bottom)+longth(b.top,b.bottom)-longth(min(a.top,b.top),max(a.bottom,b.bottom));
		if(h>0)
			return w*h；
	}
	return 0;
}

[olderma]

菜鸟写了一个，效率不咋的，不过应该易懂哈

#include<iostream>
#include<math.h>
using namespace std;
int main()
{
	double ax1,bx1,ax2,bx2,ay1,ay2,by1,by2,s;//ax1(左边), ax2（右边）, ay1（top的纵坐标）, ay2（bottom纵坐标）. 矩形B，类似，就是 bx1, bx2, by1, by2
	cin>>ax1>>ax2>>bx1>>bx2>>ay1>>ay2>>by1>>by2;
if(((ax1<=bx1)&&(bx1<=ax2))&&((ay2<=by2)&&(by2<=ay1)))
{
	s=(ax2-bx1)*(ay1-by2);
	cout<<"面积="<<s;
}
else
	if(((ax1<=bx1)&&(bx1<=ax2))&&((by2<=ay2)&&(ay2<=by1)))
	{
		s=(ax2-bx1)*(by1-ay2);
		cout<<"面积="<<s;
	}
	else
	{
		if(((bx1<=ax1)&&(ax1<=bx2))&&((ay2<=by2)&&(by2<=ay1)))
		{
			s=(bx2-ax1)*(ay1-by2);
			cout<<"面积="<<s;
		}
		else
		{
			if(((bx1<=ax1)&&(ax1<=bx2))&&((ay2<=by2)&&(by2<=ay1)))
			{
				s=(bx2-ax1)*(ay1-by2);
			    cout<<"面积="<<s;
			}
			else
			{
				cout<<"-1";
			}
		}
	}
}

[xcszbdnl]

不就是一个线段树的用法么？？

[corn8888]

汇编都上来了

[hahajluzxb]

#include <stdio.h>
main()
{

if(ax1>bx2&&ax2<bx1&&ay1<by2&&ay2>by1)
return -1;
else
return min(|bx2-ax1|,|ax2-bx1|,|bx2-bx1|,|ax2-ax1|)*min(|by2-ay1|,|ay2-by1|,|by2-by1|,|ay2-ay1|);


}
随手写的一个，不知道对不对，望指正，没写全

[windkismet]

int area(rect const& a, rect const& b)
{
...
}
ax1\ax2\ay1\ay2\bx1\bx2\by1\by2 根据a\b中的成员替换

就不填函数了，直接给思路，欢迎各位提疑指正：

新的相交矩形定义： 左上角(x1,y1)右下角(x2,y2) 有效矩形，当且仅当 x1<x2 && y1<y2

函数实现如下：
{
x1 = (ax1>bx1)?ax1:bx1;
y1 = (ay1>by1)?ay1:by1;
x2 = (ax2<bx2)?ax2:bx2;
y2 = (ay2<by2)?ay2:by2;

if(x1<x2 && y1<y2)
return ((x2-x1)*(y2-y1));

return -1;
}

[hahajluzxb]

#include <stdio.h>
#define mintwo(x,y) ((x)>(y)?(y):(x))
#define min(x,y,z,a) mintwo(mintwo(x,y),mintwo(z,a))
main()
{
int ax1,ax2,ay1,ay2,bx1,bx2,by1,by2;
scanf("%d %d %d %d",&ax1,&ax2,&ay1,&ay2);
scanf("%d %d %d %d",&bx1,&bx2,&by1,&by2);
if(!(ax1<ax2&&ay1<ay2)||!(bx1<bx2&&by1<by2))
{
printf("error,one or more rect ");
return;
}


if(ax1>bx2||ax2<bx1||ay1<by2||ay2>by1)
printf("-1\n");
else
printf("%d\n", min(bx2-ax1,ax2-bx1,bx2-bx1,ax2-ax1)*min(ay1-by2,by1-ay2,by2-by1,ay2-ay1));


}
晕，看了一下，发现写错了，重写下吧

[zsc09_leaf]

#define MAX(a,b) (a)>(b)?(a):(b)
#define MIN(a,b) (a)>(b)?(b):(a)
typedef struct rect
{
double left,top,right,bottoom;
}rect;
double area(rect const& a, rect const& b)
{
if(!(b.right > a.left && b.top > a.bottoom && b.left < a.right && b.bottoom < a.top))
{
return 0.00;
}
double h = (MIN(a.top,b.top)) - (MAX(a.bottoom,b.bottoom));
double l = (MIN(a.right,b.right))- (MAX(a.left,b.left));
return h*l;
}

[zsc09_leaf]

#define MAX(a,b) (a)>(b)?(a):(b)
#define MIN(a,b) (a)>(b)?(b):(a)

typedef struct rect
{
	double left,top,right,bottoom;
}rect;

double area(rect const& a, rect const& b)
{
	if(!(b.right > a.left && b.top > a.bottoom && b.left < a.right && b.bottoom < a.top))
	{
		return 0.00;
	}
	double h = (MIN(a.top,b.top)) - (MAX(a.bottoom,b.bottoom));
	double l = (MIN(a.right,b.right))- (MAX(a.left,b.left));
	return h*l;
}

[zsc09_leaf]

这里暂且不考虑精度问题哦。

[onlyonesai]

[Quote=引用 80 楼 的回复:]

菜鸟写了一个，效率不咋的，不过应该易懂哈C/C++ code
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
double ax1,bx1,ax2,bx2,ay1,ay2,by1,by2,s;//ax1(左边), ax2（右边）, ay1（top的纵坐标）, ay2（bottom纵坐标）. 矩形……
[/Quote]
……这个不对吧