二分。。。。
Squares
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Time Limit: 3500MS |
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Memory Limit: 65536K |
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Total Submissions: 14530 |
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Accepted: 5488 |
Description
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.
So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
Sample Output
1
6
1
Source
Rocky Mountain 2004
枚举两个点计算另外俩个，二分查找。。。。。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>
using namespace std;
struct node
{
int x,y;
}p[1100],p1,p2;
int n,cnt;
bool cmp(node a,node b)
{
if(a.x!=b.x) return a.x<b.x;
return a.y<b.y;
}
bool findp(node a)
{
int s=0,t=n-1,m;
int X=a.x,Y=a.y;
while(s<=t)
{
m=(s+t)/2;
if(p.x==X&&p.y==Y) return true;
if(p.x<X||(p.x==X&&p.y<Y)) s=m+1;
else t=m-1;
}
return false;
}
int main()
{
while(scanf("%d",&n)!=EOF&&n)
{
cnt=0;
for(int i=0;i<n;i++)
{
scanf("%d%d",&p.x,&p.y);
}
sort(p,p+n,cmp);
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
p1.x=p.x+p[j].y-p.y; p1.y=p.y+p.x-p[j].x;
p2.x=p[j].x+p[j].y-p.y; p2.y=p[j].y+p.x-p[j].x;
if(!findp(p1)) continue;
if(findp(p2)) cnt++;
}
}
printf("%d\n",cnt/2);
}
return 0;
}
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* This source code was highlighted by YcdoiT. ( style: Codeblocks )
