CodeForces 474C - Captain Marmot

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;}