Simple geometry (judging the position of the rectangle) uvalive 7070 the E-pang Palace (14 Guangzhou B)

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Test instructions: Given some points, ask to make up the area and maximum of two disjoint rectangles

Analysis: Violent enumeration, first find the two points that can make up the rectangle and save it (vis array very good), and then write a function to determine whether four points inside another rectangle. There was no save rectangle, with for to find the rectangle, the result is confused to forget to judge the situation of the back shape ...

/************************************************* author:running_time* Created TIME:2015/11/6 Friday 17:00:44* Fil E Name:B_2.cpp ************************************************/#include <cstdio> #include <algorithm># Include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string > #include <vector> #include <queue> #include <deque> #include <stack> #include <list># Include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime>using namespace std; #define Lson L, Mid, RT << 1#define Rson mid + 1, R, RT << 1 | 1typedef long ll;const int N = 2e2 + 10;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const Double EPS = 1e-10;c    Onst Double PI = ACOs ( -1.0); struct point {int x, y; Point () {}-(int x, int y): x (x), Y (y) {} bool operator < (const point &r) Const {if (x = = R . x) return Y < R. Y;    else return x < r.x;    }}p[33];struct Matrix {Point A, B; Matrix () {} matrix (point A, point B): A (a), B (b) {}};vector<matrix> Mat;bool vis[n][n];int inside (point P, p Oint A, point B) {if (p.x >= a.x && p.x <= b.x && p.y <= a.y && p.y >= b.y  {if (p.x > a.x && p.x < b.x && P.y < a.y && p.y > B.y) return        -1;    else return 1; } else return 0;} int Area_mat (int i) {return (mat[i].b.x-mat[i].a.x) * (MAT[I].A.Y-MAT[I].B.Y);}    int judge (int i, int j) {Point ics = Point (mat[i].a.x, mat[i].b.y), id = point (mat[i].b.x, MAT[I].A.Y);    int res1 = Inside (mat[i].a, MAT[J].A, mat[j].b);    int res2 = Inside (mat[i].b, MAT[J].A, mat[j].b);    int res3 = Inside (IC, MAT[J].A, mat[j].b);    int res4 = inside (ID, mat[j].a, mat[j].b);    if (!res1 &&!res2 &&!res3 &&!res4) return 0; else if(Res1 = =-1 && res2 = =-1 && res3 = =-1 && res4 = = 1) return-1; else return 1;}    int main (void) {int n;        while (scanf ("%d", &n) = = 1) {if (!n) break;        Mat.clear ();        Memset (Vis, false, sizeof (VIS));            for (int i=0; i<n; ++i) {scanf ("%d%d", &p[i].x, &AMP;P[I].Y);        VIS[P[I].X][P[I].Y] = true;        } sort (P, p+n);            for (int i=0; i<n; ++i) {int x1 = p[i].x, y1 = p[i].y;                for (int j=i+1; j<n; ++j) {int x2 = p[j].x, y2 = p[j].y;                if (x1 >= x2 | | y1 <= y2) continue;                if (!vis[x1][y2] | |!vis[x2][y1]) continue;            Mat.push_back (Matrix (Point (x1, y1), point (x2, y2)));        }} int ans = 0; for (int i=0; I<mat.size (), ++i) {for (int j=i+1; j<mat.size (); ++j) {int res1 = Jud                GE (i, j); int res2 = JUDGE (J, I);                if (!res1 &&!res2) {ans = max (ans, area_mat (i) + Area_mat (j));                } if (res1 = =-1 | | res2 = = 1) {ans = max (ans, max (Area_mat (i), Area_mat (j)));        }}} if (ans = = 0) puts ("imp");    else printf ("%d\n", ans);    }//cout << "time Elapsed:" << 1.0 * Clock ()/clocks_per_sec << "s.\n"; return 0;}

　

This is a very lame code to stay here and read ...

/************************************************* author:running_time* Created TIME:2015/10/14 Wednesday 14:59:33* Fi Le name:b.cpp ************************************************/#include <cstdio> #include <algorithm># Include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string > #include <vector> #include <queue> #include <deque> #include <stack> #include <list># Include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime>using    namespace Std;const int N = 33;const int M = 210;const int INF = 0x3f3f3f3f;struct point {int x, y;        BOOL operator < (const point &r) Const {if (x = = r.x) return y < r.y;    else return x < r.x;    }}p[n];vector<int> Vx[m], vy[m];int cal (int la, int lb) {if (La < 0) La =-la;    if (lb < 0) lb =-lb; return La lb;    int main (void) {int n; while (sCANF ("%d", &n) = = 1) {if (!n) break;        for (int i=0; i<=200; ++i) {vx[i].clear (); Vy[i].clear ();        } for (int i=1; i<=n; ++i) {scanf ("%d%d", &p[i].x, &AMP;P[I].Y);        } sort (p+1, p+1+n);   for (int i=1; i<=n; ++i) {vx[p[i].x].push_back (i);        Vy[p[i].y].push_back (i);        } int ans = 0;        BOOL flag = FALSE;            for (int i=1; i<=n; ++i) {//i one int x = p[i].x, y = p[i].y;            if (Vx[x].size () <= 1 | | vy[y].size () <= 1) continue;                for (int j=0; j<vx[x].size (); ++j) {//j, int JJ = Vx[x][j];                if (JJ = = I | | p[jj].y <= y) continue;                int JY = P[JJ].Y;                    for (int k=0; k<vy[y].size (); ++k) {//k three int kk = vy[y][k];                    if (KK = = I | | p[kk].x <= x) continue; int kx = P[kk].x;                    if (Vx[kx].size () <= 1) continue;                        for (int l=0; L<vx[kx].size (), ++l) {//l four int ll = vx[kx][l];                        if (ll = = KK | | p[ll].y <= y) continue;                                                        if (p[ll].y = JY) {//find the first rectangle                                    if (Vy[jy].size () >= 4) {for (int r=0; r<vy[jy].size (); ++r) {                                    int rr = Vy[jy][r];                                    if (rr = = LL | | rr = = JJ) continue;                                        for (int r3=0; r3<vy[jy].size (); ++r3) {int r4 = VY[JY][R3];                                        if (R4 = = LL | | r4 = = JJ | | r4 = = RR) continue;                                        int xx = p[rr].x, kkx = p[r4].x; if (x <= xX && xx <= kx) continue;                                        if (x <= kkx && kkx <= kx) Continue;                                        if (Vx[xx].size () <= 1 | | vx[kkx].size () <= 1) continue;                                            for (int o=0; o<vx[xx].size (); ++o) {int oo = Vx[xx][o];                                            if (oo = = RR | | p[oo].y <= JY) continue;                                            int jjy = P[OO].Y;                                                for (int u=0; u<vx[kkx].size (); ++u) {int uu = Vx[kkx][u];                                                if (UU = = R4 | | p[uu].y! = JJY) Continue;                                                Flag = true;                                            ans = max (ans, Cal (Kx-x, jy-y) + cal (Kkx-xx, JJY-JY));    }                                    }                                    }                                } } for (int yy=jy+1; yy<=200; ++yy) {if (vy[y                                Y].size () <= 1) continue;                                    for (int yi=0; yi<vy[yy].size (); ++yi) {int II = Vy[yy][yi];     int xx = p[ii].x, yy = P[ii].y;                                        To find the second rectangle for (int j3=0; j3<vx[xx].size (); ++j3) {                                        int J4 = vx[xx][j3];                                        if (J4 = = II | | p[j4].y <= yy) continue;                                        int jjy = P[J4].Y;                                            for (int k3=0; k3<vy[yy].size (); ++k3) {int K4 = VY[YY][K3];  if (K4 = = II)Continue                                            int kkx = p[k4].x;                                                for (int l3=0; l3<vx[kkx].size (); ++l3) {int L4 = Vx[kkx][l3];                                                if (L4 = = K4 | | p[l4].y! = JJY) Continue;                                                Flag = true;                                            ans = max (ans, Cal (Kx-x, jy-y) + cal (Kkx-xx, jjy-yy));                            }                                        }                                    }                                }                                } if (Vx[kx].size () >= 4) {                                    for (int r=0; r<vx[kx].size (); ++r) {int rr = vx[kx][r];                           if (rr = = KK | | rr = = LL) continue;         for (int r3=0; r3<vy[jy].size (); ++r3) {int r4 = VX[KX][R3];                                        if (R4 = = KK | | r4 = LL | | r4 = = RR) continue;                                        int yy = p[rr].y, jjy = p[r4].y;                                        if (yy > Jjy) Swap (yy, JJY);                                        if (y <= yy && yy <= jy) continue;                                        if (y <= jjy && jjy <= JY) continue;                                        if (Vy[yy].size () <= 1 | | vy[jjy].size () <= 1) continue;                                            for (int o=0; o<vy[yy].size (); ++o) {int oo = Vy[yy][o];                                            if (oo = = RR | | p[oo].x <= KX) Continue;                                            int kkx = p[oo].x; for (int u=0; u<vy[Jjy].size ();                                                ++u) {int uu = Vy[jjy][u];                                                if (UU = = R4 | | p[uu].x! = KKX) Continue;                                                Flag = true;                                            ans = max (ans, Cal (Kx-x, jy-y) + cal (Kkx-kx, JJY-JY));                            }                                        }                                    }                                } } for (int x5=kx+1; x5<=200; ++x5) {if (                                Vx[x5].size () <= 1) continue;                                    for (int xi=0; xi<vx[x5].size (); ++xi) {int II = VX[X5][XI];     int xx = p[ii].x, yy = P[ii].y;          To find the second rectangle for (int j3=0; j3<vx[x5].size (); ++j3) {                              int JJ = Vx[x5][j3];                                        if (JJ = = II | | p[jj].y <= yy) continue;                                        int jjy = P[JJ].Y;                                        if (Vy[yy].size () <= 1) continue;                                            for (int k3=0; k3<vy[yy].size (); ++k3) {int K4 = VY[YY][K3];                                            if (K4 = = II | | p[k4].x <= XX) continue;                                            int kkx = p[k4].x;                                                for (int l3=0; l3<vx[kkx].size (); ++l3) {int L4 = Vx[kkx][l3];                                                if (L4 = = K4 | | p[l4].y! = JJY) Continue;                                                Flag = true;      ans = max (ans, Cal (Kx-x, jy-y) + cal (Kkx-xx, jjy-yy));                                      }                                        }                                    }            }                            }                        }                    }                }        }} if (flag) {printf ("%d\n", ans);    } else puts ("imp"); } return 0;}

　　

Simple geometry (judging the position of the rectangle) uvalive 7070 the E-pang Palace (14 Guangzhou B)