CodeForces 474C - Captain Marmot（暴力）

题意：每组共4个点，4个点分别有起始坐标和轴心坐标，每个点每次都可以绕自己的轴心坐标90°，问能否用最少次数内4个点组成一个正方形，若能则输出最少次数，否则输出-1。

每个点有4个方位可以选择，共4个点，4 * 4 * 4 * 4 = 256种，暴力即可

 

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
typedef long long ll;  
typedef unsigned long long llu;  
const int MAXN = 100 + 10;  
const int MAXT = 10000 + 10;  
const int INF = 0x7f7f7f7f;  
const double pi = acos(-1.0);  
const double eps = 1e-6;  
using namespace std;  
  
struct Point{  
    int x, y;  
    bool operator < (const Point &b) const{  
        return x < b.x || (y < b.y && x == b.x);  
    }  
    Point(int a = 0, int b = 0) : x(a), y(b) {}  
    Point(const Point &b) : x(b.x), y(b.y) {}  
    Point& operator = (const Point &b){  
        x = b.x;  y = b.y;  
        return *this;  
    }  
}p[5];  
  
int xx[5], yy[5], a[5], b[5];  
  
Point f1(int x, int y, int ox, int oy){  
    return Point(x, y);  
}  
  
Point f2(int x, int y, int ox, int oy){  
    return Point(ox + oy - y, oy - ox + x);  
}  
  
Point f3(int x, int y, int ox, int oy){  
    return Point(2 * ox - x, 2 * oy - y);  
}  
  
Point f4(int x, int y, int ox, int oy){  
    return Point(y - oy + ox, ox - x + oy);  
}  
  
bool judge(){  
    sort(p, p + 4);  
    int c1x = p[1].x - p[0].x;  
    int c1y = p[1].y - p[0].y;  
    int c1 = c1x * c1x + c1y * c1y;  
  
    int c2x = p[2].x - p[0].x;  
    int c2y = p[2].y - p[0].y;  
    int c2 = c2x * c2x + c2y * c2y;  
  
    int c3x = p[3].x - p[2].x;  
    int c3y = p[3].y - p[2].y;  
    int c3 = c3x * c3x + c3y * c3y;  
  
    int c4x = p[3].x - p[1].x;  
    int c4y = p[3].y - p[1].y;  
    int c4 = c4x * c4x + c4y * c4y;  
  
    if(c1 == c2 && c2 == c3 && c3 == c4){  
        if(c1x * c2x + c1y * c2y == 0 && c2x * c3x + c2y * c3y == 0 && c3x * c4x + c3y * c4y == 0 && (p[0].x != p[1].x || p[0].y != p[1].y))  
            return true;  
    }  
    return false;  
}  
  
int main(){  
    int T;  
    scanf("%d", &T);  
    while(T--){  
        for(int i = 0; i < 4; ++i)  scanf("%d%d%d%d", &xx[i], &yy[i], &a[i], &b[i]);  
        int ans = INF;  
        for(int i = 0; i < 4; ++i)  
        for(int j = 0; j < 4; ++j)  
        for(int k = 0; k < 4; ++k)  
        for(int l = 0; l < 4; ++l){  
            if(i == 0)  p[0] = f1(xx[0], yy[0], a[0], b[0]);  
            else if(i == 1)  p[0] = f2(xx[0], yy[0], a[0], b[0]);  
            else if(i == 2)  p[0] = f3(xx[0], yy[0], a[0], b[0]);  
            else if(i == 3)  p[0] = f4(xx[0], yy[0], a[0], b[0]);  
  
            if(j == 0)  p[1] = f1(xx[1], yy[1], a[1], b[1]);  
            else if(j == 1)  p[1] = f2(xx[1], yy[1], a[1], b[1]);  
            else if(j == 2)  p[1] = f3(xx[1], yy[1], a[1], b[1]);  
            else if(j == 3)  p[1] = f4(xx[1], yy[1], a[1], b[1]);  
  
            if(k == 0)  p[2] = f1(xx[2], yy[2], a[2], b[2]);  
            else if(k == 1)  p[2] = f2(xx[2], yy[2], a[2], b[2]);  
            else if(k == 2)  p[2] = f3(xx[2], yy[2], a[2], b[2]);  
            else if(k == 3)  p[2] = f4(xx[2], yy[2], a[2], b[2]);  
  
            if(l == 0)  p[3] = f1(xx[3], yy[3], a[3], b[3]);  
            else if(l == 1)  p[3] = f2(xx[3], yy[3], a[3], b[3]);  
            else if(l == 2)  p[3] = f3(xx[3], yy[3], a[3], b[3]);  
            else if(l == 3)  p[3] = f4(xx[3], yy[3], a[3], b[3]);  
  
            if(judge())  ans = min(ans, i + j + k + l);  
        }  
        if(ans == INF)  printf("-1\n");  
        else  printf("%d\n", ans);  
    }  
    return 0;  
}