UVa 11123 - Counting Trapizoid

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

給平面上 N 個不重複的點,有多少梯形可以由這幾個點構成?

  • 有四條邊
  • 一對平行、一對不平行
  • 面積為正

Sample Input

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4
0 0
1 0
0 1
1 1
4
0 0
1 1
2 0
0 1
0

Sample Output

1
2
Case 1: 0
Case 2: 1

Solution

先窮舉任兩個點拉起的線段 (let N = n*(n-1)/2)。

針對線段斜率排序,接著將相同斜率放置在同一個 group。

對於每一個 group 的每一個元素計算有多少可以配對成梯形。

  • 防止共線
  • 防止線段具有相同長度 (防止兩對邊平行)

斜率相同,按照交 y 軸的大小由小到大排序。(左側到右側),由左側掃描到右側,依序把非共線的線段丟入,並且紀錄線段長度資訊。每一個能匹配的組合為 當前線段數 - 相同長度線段數

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#include <stdio.h> 
#include <math.h>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <assert.h>
using namespace std;

#define eps 1e-9
struct Pt {
double x, y;
Pt(double a = 0, double b = 0):
x(a), y(b) {}
Pt operator-(const Pt &a) const {
return Pt(x - a.x, y - a.y);
}
Pt operator+(const Pt &a) const {
return Pt(x + a.x, y + a.y);
}
bool operator<(const Pt &a) const {
if (fabs(x - a.x) > eps)
return x < a.x;
if (fabs(y - a.y) > eps)
return y < a.y;
return false;
}
};
double dist(Pt a, Pt b) {
return hypot(a.x - b.x, a.y - b.y);
}
double dot(Pt a, Pt b) {
return a.x * b.x + a.y * b.y;
}
double cross(Pt o, Pt a, Pt b) {
return (a.x-o.x)*(b.y-o.y)-(a.y-o.y)*(b.x-o.x);
}
int between(Pt a, Pt b, Pt c) {
return dot(c - a, b - a) >= -eps && dot(c - b, a - b) >= -eps;
}
int onSeg(Pt a, Pt b, Pt c) {
return between(a, b, c) && fabs(cross(a, b, c)) < eps;
}
struct Seg {
Pt s, e;
double angle;
int label;
Seg(Pt a = Pt(), Pt b = Pt(), int l=0):s(a), e(b), label(l) {
angle = atan2(e.y - s.y, e.x - s.x);
}
bool operator<(const Seg &other) const {
if (fabs(angle - other.angle) > eps)
return angle > other.angle;
if (cross(other.s, other.e, s) > eps)
return true;
return false;
}
};
struct DOUBLE {
double v;
DOUBLE(double a = 0):
v(a) {}
bool operator<(const DOUBLE &other) const {
if (fabs(v - other.v) > eps)
return v < other.v;
return false;
}
};
Pt D[256];
Seg segs[65536], buf[65536];

int main() {
int cases = 0;
int n, m;
while (scanf("%d", &n) == 1 && n) {
for (int i = 0; i < n; i++)
scanf("%lf %lf", &D[i].x, &D[i].y);
sort(D, D + n);

m = 0;
for (int i = 0; i < n; i++)
for (int j = i+1; j < n; j++)
segs[m++] = Seg(D[i], D[j]);

sort(segs, segs + m);
long long ret = 0;
for (int i = 0; i < m; ) {
int j = i, tn = 0;
while (j < m && fabs(segs[i].angle - segs[j].angle) < eps)
buf[tn++] = segs[j], j++;
i = j;

map<DOUBLE, int> LEN; // LEN[len] = count.
queue<Seg> Q;
int tmp = 0;
// for (int k = 0; k < tn; k++)
// printf("(%lf %lf) - (%lf %lf)\n", buf[k].s.x, buf[k].s.y, buf[k].e.x, buf[k].e.y);
// puts("--- same slope");
for (int k = 0; k < tn; k++) {
while (!Q.empty() && fabs(cross(buf[k].s, buf[k].e, Q.front().s)) > eps) {
Seg t = Q.front();
Q.pop();
tmp++, LEN[DOUBLE(dist(t.s, t.e))]++;
}
int comb = tmp - LEN[DOUBLE(dist(buf[k].s, buf[k].e))]; // remove same length
ret += comb;
Q.push(buf[k]);
}
}
printf("Case %d: %lld\n", ++cases, ret);
}
return 0;
}
/*
4
0 0
1 0
0 1
1 1

4
0 0
1 1
2 0
0 1

6
5 1
3 1
5 0
3 3
2 4
0 8



0
*/