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数论:zoj 2540 Form a Square

 
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【转】http://blog.csdn.net/zxy_snow/article/details/6153629

 

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
using namespace std;
typedef struct{
	double x,y;
}coor;
coor c[4];
double len(double x,double y,double xx,double yy)
{
	return (x-xx)*(x-xx)+(y-yy)*(y-yy);
}
int compute(coor a,coor b,coor c,coor d)
{
	if( len(a.x,a.y,c.x,c.y) == len(a.x,a.y,b.x,b.y) + len(b.x,b.y,c.x,c.y) )
		if( len(b.x,b.y,c.x,c.y) + len(c.x,c.y,d.x,d.y) == len(b.x,b.y,d.x,d.y) )
			if( len(a.x,a.y,b.x,b.y) == len(b.x,b.y,c.x,c.y) )
				return 1;
	return 0;
}
int main()
{
	int ncases,ind = 1,blank = 0,i,j,k,x,flag;
	scanf("%d",&ncases);
	while( ncases-- )
	{
		flag = 0;
		if( blank ) printf("/n");
		blank = 1;
		for(i=0; i<4; i++)
			scanf("%lf %lf",&c[i].x,&c[i].y);
		for(i=0; i<4; i++)
			for(j=0; j<4; j++)
			{
				if( j == i ) continue;
				for(k=0; k<4; k++)
				{
					if( k == i || k == j ) continue;
					x = 6 - k - i - j;
					if( compute(c[i],c[j],c[k],c[x]) == 1 )
					{
						flag = 1;
						goto end;
					}
				}
			}
		end:;
		printf("Case %d:/n",ind++);
		if( flag ) 
			printf("Yes/n");
		else 
			printf("No/n");
	}
return 0;
}

 

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