UVA 11123 - Counting Trapizoid(计数问题+容斥)

UVA 11123 - Counting Trapizoid

题目链接

题意：给定一些点，不重复，求出一共有几个梯形

思路：先把所有两点组成直线求出来，然后排序，斜率相同的C2n个，然后再扣除掉重叠的直线情况和长度相等情况（这样为平行四边形或矩形），由于扣除的时候会重复扣掉重叠和相等，所以在加回来，这是容斥原理。

代码：

#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
;

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);
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;
}