King's Sanctuary（简单几何）

King's Sanctuary

Time Limit: 1000 ms Memory Limit: 65535 kB Solved: 111 Tried: 840

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Description

The king found his adherents were building four sanctuaries for him. He is interested about the positions of the sanctuaries and wants to know whether they would form a parallelogram, rectangle, diamond, square or anything else.

Input

The first line of the input is T(1 <= T <= 1000), which stands for the number of test cases you need to solve.
Each case contains four lines, and there are two integers in each line, which shows the position of the four sanctuaries. And it is guaranteed that the positions are given clockwise. And it is always a convex polygon, if you connect the four points clockwise.

Output

For every test case, you should output "Case #t: " first, where t indicates the case number and counts from 1, then output the type of the quadrilateral.

Sample Input

5
0 0
1 1
2 1
1 0
0 0
0 1
2 1
2 0
0 0
2 1
4 0
2 -1
0 0
0 1
1 1
1 0
0 0
1 1
2 1
3 0

Sample Output

Case #1: Parallelogram
Case #2: Rectangle
Case #3: Diamond
Case #4: Square
Case #5: Others

Source

Sichuan State Programming Contest 2012

步骤：1.判断是否平行四边形，不是则为others，是则执行2；2.判断是否菱形，是否长方形，都是则为正方形，都不是则为平行四边形，其一是另一否，则是菱形或长方形。

#include <iostream>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

int T, x[4], y[4];

bool Parallelogram()
{
int a = (y[1] - y[0]) * (x[3] - x[2]);
int b = (x[1] - x[0]) * (y[3] - y[2]);
if(a != b) return false;
a = (x[2] - x[1]) * (y[3] - y[0]);
b = (x[3] - x[0]) * (y[2] - y[1]);
return a == b;
}

bool Rectangle()
{
int a = (x[0] - x[2]) * (x[0] - x[2]) + (y[0] - y[2]) * (y[0] - y[2]);
int b = (x[1] - x[3]) * (x[1] - x[3]) + (y[1] - y[3]) * (y[1] - y[3]);
return a == b;
}

bool Diamond()
{
int a = (y[0] - y[2]) * (y[1] - y[3]);
int b = (x[1] - x[3]) * (x[0] - x[2]);
return a == -b;
}

int main()
{
scanf("%d", &T);
for(int ca = 1; ca <= T; ca++)
{
int i, j;
for(i = 0; i < 4; i++)
{
scanf("%d %d", &x[i], &y[i]);
}
printf("Case #%d: ", ca);
for(i = 0; i < 4; i++)
{
for(j = i + 1; j < 4; j++)
{
if(x[i] == x[j] && y[i] == y[j])
break;
}
if(j != 4) break;
}
if(i != 4) puts("Others");
else
{
bool tag, tag1;
tag = Parallelogram();
if(tag == false) {puts("Others");}
else
{
tag = Rectangle();
tag1 = Diamond();
if(tag == false && tag1 == false) puts("Parallelogram");
else if(tag == true && tag1 == true) puts("Square");
else if(tag == true) puts("Rectangle");
else if(tag1 == true) puts("Diamond");
}
}
}
return 0;
}