UVA 11123-counting trapizoid (counting problem + tolerance)

UVA 11123-counting trapizoid

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Test instructions: Given some points, do not repeat, find a total of a few trapezoidal

Idea: First put all two points into a straight line to find out, and then sort, slope the same c 2 n , and then deduct the overlapping straight line condition and the length equal case (this is parallelogram or rectangle), because the deduction will be repeated deduction of overlap and equality, so in addition, this is the principle of repulsion.

Code:

#include <stdio.h> #include <string.h> #include <math.h> #include <algorithm>using namespace Std;const double eps = 1e-9;const double pi = ACOs ( -1.0); const int N = 205;int N, ln;struct point {double X, y;} P[n]; struct Line {Double L, a, B, C, K, y;} l[n * N];bool CMPK (line A, line B) {return A.K < B.K;} BOOL Cmpl (line A, line B) {return A.L < B.L;}    BOOL Cmpy (line A, line B) {if (Fabs (A.Y-B.Y) < EPS) return A.L < B.L; return A.Y < B.y;}    Line build (Point A, point B) {line ans;    ANS.L = sqrt ((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y));    ANS.A = b.y-a.y, ans.b = a.x-b.x, ans.c = ANS.A * a.x + ans.b * A.Y;    ANS.K = atan2 (ANS.A,-ans.b);    if (ANS.K < 0 | | fabs (ANS.K) < EPS) ANS.K + = PI;    if (Fabs (ans.b) < EPS) ans.y = ANS.C/ANS.A;    else Ans.y = ans.c/ans.b; return ans;} Long Long C (long long N) {return n * (n-1)/2;}    Long long Solve () {sort (L, l + ln, CMPK); A long long ans = 0,CNT = 0;    Line Save[n * N];    int sn = 0;    L[LN++].K =-1;    CNT = 0;    for (int i = 0; i < ln; i++) {if (!i | | fabs (L[I].K-L[I-1].K) < EPS) {save[sn++] = L[i];    cnt++; Continue;} Ans + = C (CNT), CNT = 0;sort (save, Save + sn, Cmpl), for (int j = 0; J < SN; j + +) {if (!j | | fabs (save[j-1].l-save    [J].L) < EPS) {cnt++;continue;    } ans-= C (CNT); CNT = 1;} Ans-= C (CNT), CNT = 0;sort (save, Save + sn, cmpy), for (int j = 0; J < SN; j + +) {if (!j | | fabs (save[j-1].y-save    [J].y) < EPS) {cnt++;continue;    } ans-= C (CNT); CNT = 1;} Ans-= C (CNT); cnt = 0;for (int j = 0; J < SN; j + +) {if (!j | | (Fabs (SAVE[J-1].Y-SAVE[J].Y) < EPS && fabs (SAVE[J-1].L-SAVE[J].L) < EPS))    {cnt++;continue;    } ans + = C (CNT); CNT = 1;}    Ans + = C (CNT); sn = 0;save[sn++] = l[i];cnt = 1; } return ans;    int main () {int cas = 0; while (~SCANF ("%d", &n) && n) {ln = 0;for (int i = 0; i < n; i++) scanf ("%lf%lF ", &p[i].x, &p[i].y); for (int i = 0; i < n; i++) {for (int j = i + 1; j < N; j + +) {l[ln++] = build (P[i]    , P[j]);    }}printf ("Case%d:%lld\n", ++cas, Solve ()); } return 0;}