计算几何初步
计算几何基础
1. 判断点是否在线段上
叉积必为 0 保证在延长线上，点积不大于 0 保证不会在线段的两侧.
int Onsegment(point tmp, point a, point b) {
if(dcmp(cross(a - tmp, b - tmp)) == 0 && dcmp(dot(a - tmp, b - tmp)) <= 0)
return 1;
return 0;
}
2. 判断两个线段是否有相交（不在端点处）
看 $\mathrm{(a,b)}$ 和 $\mathrm{(c,d)}$ 分别都在各自两侧.
// 前面是线段上，后面是不在线段上相交.
int Line_Intersect(point a, point b, point c, point d) {
double x1 = cross(b - a, c - a), y1 = cross(b - a, d - a);
double x2 = cross(d - c, a - c), y2 = cross(d - c, b - c);
if(!dcmp(x1) || !dcmp(x2) || !dcmp(y1) || !dcmp(y2)) {
bool f1 = Onsegment(a, c, d);
bool f2 = Onsegment(b, c, d);
bool f3 = Onsegment(c, a, b);
bool f4 = Onsegment(d, a, b);
bool f = (f1 | f2 | f3 | f4);
return f;
}
if(dcmp(x1) * dcmp(y1) < 0 && dcmp(x2) * dcmp(y2) < 0)
return 1;
return 0;
}
3. 凸包
$\mathrm{Graham}$ 扫描法：
将 $\mathrm{y}$ 坐标最小的点作为基准给其他点按照角度逆时针排序.
加入新点的时候看之前加的边是否在新加的边的左侧，如果是则弹掉.
// b 在 a 的逆时针为正，夹角为 a 转到 b 的有向角度（sin)
double cross(Vector a, Vector b) {
return a.x * b.y - a.y * b.x;
}
int n ;
point p[N], A[N];
bool cmp2(point i, point j) {
int t = dcmp(cross(i - p[1], j - p[1]));
if(t != 0)
return t > 0;
else {
return dis(i, p[1]) < dis(j, p[1]);
}
}
int main() {
// setIO("input");
scanf("%d", &n);
for(int i = 1; i <= n ; ++ i) {
scanf("%lf%lf", &p[i].x, &p[i].y);
// if(p[i].y < p[1].y) swap(p[i], p[1]);
}
sort(p + 1, p + 1 + n);
sort(p + 2, p + 1 + n, cmp2);
int cnt = 1;
A[1] = p[1];
for(int i = 2; i <= n ; ++ i) {
while(cnt > 1 && dcmp(cross(A[cnt] - A[cnt - 1], p[i] - A[cnt])) != 1)
-- cnt ;
A[++ cnt] = p[i];
}
A[++ cnt] = p[1];
double ans = 0.0;
for(int i = 1; i < cnt ; ++ i)
ans += dis(A[i], A[i + 1]);
printf("%.2f", ans);
return 0;
}
4. 线段和直线是否有交点
int seg_inter_line(Line l1, Line l2) {
if(dcmp(cross(l1.e - l2.s, l1.s - l2.s)) * dcmp(cross(l1.e - l2.e, l1.s - l2.e)) <= 0)
return 1;
return 0;
}
5. 直线和直线的交点
// l1 为直线，l2 为线段, 看是否相交
int seg_inter_line(Line l1, Line l2) {
if(dcmp(cross(l1.e - l2.s, l1.s - l2.s)) * dcmp(cross(l1.e - l2.e, l1.s - l2.e)) <= 0)
return 1;
return 0;
}
// 看直线 l1 和 l2 的位置情况.
int judge_Line(Line l1, Line l2) {
if(dcmp(cross(l1.e - l1.s, l2.e - l2.s)) == 0) {
// 0 : 共线，-1：平行.
if(seg_inter_line(l1, l2) == 1)
return 0;
else
return -1;
}
return 1;
}
// 求直线 l1 与 l2 的交点
point Get_Line(Line l1, Line l2) {
Vector v1 = l1.e - l1.s, v2 = l2.e - l2.s;
double t = cross(l2.s - l1.s, v2) / cross(v1, v2);
return l1.s + v1 * t;
}
6. 判断点是否在多边形内部
// 判断点 p 是否在多边形内.
// -1: 在边界.
// 0 : 在外部
// 1 : 在内部
int ispoly(point p, point poly[], int len) {
int cnt = 0;
Line ray, side;
ray.s = p;
ray.e = point(-1000000000.0, p.y);
for(int i = 0; i < len; ++ i) {
side = Line(poly[i], poly[(i + 1) % len]);
// 在边界
if(Onsegment(p, side.e, side.s))
return -1;
// 平行线不用考虑
if(dcmp(side.s.y - side.e.y) == 0)
continue;
if(Onsegment(side.s, ray.s, ray.e)) {
// 凸包点在射线上.
if(dcmp(side.s.y - side.e.y) > 0) ++ cnt;
}
else if(Onsegment(side.e, ray.s, ray.e)) {
if(dcmp(side.e.y - side.s.y) > 0) ++ cnt;
}
else if(Line_Intersect(side.e, side.s, ray.s, ray.e) == 1) ++ cnt;
}
if(cnt & 1) return 1;
return 0;
}
TOYS
来源：POJ - 2318
对于每个点，二分在最右侧的哪条线段的右侧，判断左右用叉积.
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#define point Vector
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
const double eps = 1e-8;
struct Vector {
double x, y;
Vector(double X = 0.0, double Y = 0.0){
x = X, y = Y;
}
};
int dcmp(double x) {
if(fabs(x) < eps)
return 0;
return x < 0 ? -1 : 1;
}
bool operator < (const point &a, const point & b) {
if(dcmp(a.x - b.x) == 0)
return a.y < b.y;
else return a.x < b.x;
}
bool operator == (const point &a, const point & b) {
if(dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0)
return true;
return false;
}
Vector operator + (Vector a, Vector b) { return Vector(a.x + b.x, a.y + b.y); }
Vector operator - (Vector a, Vector b) { return Vector(a.x - b.x, a.y - b.y); }
Vector operator * (Vector a, double p) { return Vector(a.x * p, a.y * p); }
Vector operator / (Vector a, double p) { return Vector(a.x / p, a.y / p); }
double dot(Vector a, Vector b) {
return a.x * b.x + a.y * b.y;
}
double len(point a) {
return sqrt(dot(a, a));
}
double sqr(double x) {
return x * x;
}
double dis(point a, point b) {
return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}
// b 在 a 的逆时针为正，夹角为 a 转到 b 的有向角度（sin)
double cross(Vector a, Vector b) {
return a.x * b.y - a.y * b.x;
}
// 求点积再除以模长.
double Angle(Vector a, Vector b) {
return acos(dot(a, b) / len(a) / len(b));
}
// 求法向量.
Vector normal(Vector a) {
double u = len(a);
return Vector(-a.y / u, a.x / u);
}
// 看 tmp 是否在 ab 上
int Onsegment(point tmp, point a, point b) {
if(dcmp(cross(a - tmp, b - tmp)) == 0 && dcmp(dot(a - tmp, b - tmp)) <= 0)
return 1;
return 0;
}
// 前面是线段上，后面是不在线段上相交.
int Line_Intersect(point a, point b, point c, point d) {
double x1 = cross(b - a, c - a), y1 = cross(b - a, d - a);
double x2 = cross(d - c, a - c), y2 = cross(d - c, b - c);
if(!dcmp(x1) || !dcmp(x2) || !dcmp(y1) || !dcmp(y2)) {
bool f1 = Onsegment(a, c, d);
bool f2 = Onsegment(b, c, d);
bool f3 = Onsegment(c, a, b);
bool f4 = Onsegment(d, a, b);
bool f = (f1 | f2 | f3 | f4);
return f;
}
if(dcmp(x1) * dcmp(y1) < 0 && dcmp(x2) * dcmp(y2) < 0)
return 1;
return 0;
}
#define N 20000
int n, m;
struct Line {
point a, b;
}A[N];
point tp[N];
int fin[N];
int jud(int x, int i) {
// 看第 x 线段.
if(dcmp(cross(A[x].b - A[x].a, tp[i] - A[x].a)) == 1)
return 1;
return 0;
}
int main() {
// setIO("input");
while(1) {
scanf("%d", &n);
if(n == 0) break;
scanf("%d", &m);
double x[2], y[2];
scanf("%lf%lf%lf%lf", &x[0], &y[0], &x[1], &y[1]);
for(int i = 0; i <= n ; ++ i)
fin[i] = 0;
for(int i = 1; i <= n ; ++ i) {
double u, l;
scanf("%lf%lf", &u, &l);
A[i].a = point(u, y[0]);
A[i].b = point(l, y[1]);
}
for(int i = 1; i <= m ; ++ i) {
scanf("%lf%lf", &tp[i].x, &tp[i].y);
}
for(int i = 1; i <= m ; ++ i) {
int l = 1, r = n, an = 0;
while(l <= r) {
int mid = (l + r) >> 1;
if(jud(mid, i)) an = mid, l = mid + 1;
else r = mid - 1;
}
// an 的右侧，则
fin[an] ++ ;
}
for(int i = 0; i <= n ; ++ i) {
printf("%d: %d\n", i, fin[i]);
}
printf("\n");
}
return 0;
}
Segments
来源：POJ - 3304
考虑通过公共投影点的线经过旋转后一定经过所有线段，即枚举该直线两个端点然后判断是否有交.
这里就要判断一条直线和线段是否有交点了.
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#define point Vector
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
const double eps = 1e-8;
struct Vector {
double x, y;
Vector(double X = 0.0, double Y = 0.0){
x = X, y = Y;
}
};
struct Line {
point s, e;
Line() {};
Line(point i, point j) {
s = i, e = j;
}
};
int dcmp(double x) {
if(fabs(x) < eps)
return 0;
return x < 0 ? -1 : 1;
}
bool operator < (const point &a, const point & b) {
if(dcmp(a.x - b.x) == 0)
return a.y < b.y;
else return a.x < b.x;
}
bool operator == (const point &a, const point & b) {
if(dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0)
return true;
return false;
}
Vector operator + (Vector a, Vector b) { return Vector(a.x + b.x, a.y + b.y); }
Vector operator - (Vector a, Vector b) { return Vector(a.x - b.x, a.y - b.y); }
Vector operator * (Vector a, double p) { return Vector(a.x * p, a.y * p); }
Vector operator / (Vector a, double p) { return Vector(a.x / p, a.y / p); }
double dot(Vector a, Vector b) {
return a.x * b.x + a.y * b.y;
}
double len(point a) {
return sqrt(dot(a, a));
}
double sqr(double x) {
return x * x;
}
double dis(point a, point b) {
return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}
// b 在 a 的逆时针为正，夹角为 a 转到 b 的有向角度（sin)
double cross(Vector a, Vector b) {
return a.x * b.y - a.y * b.x;
}
// 求点积再除以模长.
double Angle(Vector a, Vector b) {
return acos(dot(a, b) / len(a) / len(b));
}
// 求法向量.
Vector normal(Vector a) {
double u = len(a);
return Vector(-a.y / u, a.x / u);
}
// 看 tmp 是否在 ab 上
int Onsegment(point tmp, point a, point b) {
if(dcmp(cross(a - tmp, b - tmp)) == 0 && dcmp(dot(a - tmp, b - tmp)) <= 0)
return 1;
return 0;
}
// 前面是线段上，后面是不在线段上相交.
int Line_Intersect(point a, point b, point c, point d) {
double x1 = cross(b - a, c - a), y1 = cross(b - a, d - a);
double x2 = cross(d - c, a - c), y2 = cross(d - c, b - c);
if(!dcmp(x1) || !dcmp(x2) || !dcmp(y1) || !dcmp(y2)) {
bool f1 = Onsegment(a, c, d);
bool f2 = Onsegment(b, c, d);
bool f3 = Onsegment(c, a, b);
bool f4 = Onsegment(d, a, b);
bool f = (f1 | f2 | f3 | f4);
return f;
}
if(dcmp(x1) * dcmp(y1) < 0 && dcmp(x2) * dcmp(y2) < 0)
return 1;
return 0;
}
// l1 为直线，l2 为线段
int seg_inter_line(Line l1, Line l2) {
if(dcmp(cross(l1.e - l2.s, l1.s - l2.s)) * dcmp(cross(l1.e - l2.e, l1.s - l2.e)) <= 0)
return 1;
return 0;
}
#define N 3005
Line arr[N];
int n ;
int check(point x, point y) {
if(dcmp(dis(x, y)) == 0)
return 0;
Line now = Line(x, y);
for(int i = 1; i <= n ; ++ i) {
if(!seg_inter_line(now, arr[i]))
return 0;
}
return 1;
}
void solve() {
scanf("%d", &n);
for(int i = 1; i <= n ; ++ i) {
scanf("%lf%lf", &arr[i].s.x, &arr[i].s.y);
scanf("%lf%lf", &arr[i].e.x, &arr[i].e.y);
}
int flag = 0;
for(int i = 1; i <= n ; ++ i) {
for(int j = i; j <= n ; ++ j) {
if(check(arr[i].s, arr[j].s) || check(arr[i].s, arr[j].e) || check(arr[i].e, arr[j].s) || check(arr[i].e, arr[j].e))
flag = 1;
}
if(flag) break ;
}
if(flag) {
printf("Yes!\n");
}
else {
printf("No!\n");
}
}
int main() {
// setIO("input");
int T;
scanf("%d", &T);
while(T -- ) solve();
return 0;
}
The Doors
来源：uva393
如果连边一定是在门的端点处连线.
所以问题就转化成判断所有线段是否会和一条线段相交，用模板判定后跑最短路即可.
这里注意为了严谨还要判断是否有两个线段重合的情况
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <queue>
#define point Vector
#define ll long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
const double eps = 1e-7;
struct Vector {
double x, y;
Vector(double X = 0.0, double Y = 0.0){
x = X, y = Y;
}
};
struct Line {
point s, e;
Line() {};
Line(point i, point j) {
s = i, e = j;
}
};
int dcmp(double x) {
if(fabs(x) < eps)
return 0;
return x < 0 ? -1 : 1;
}
bool operator < (const point &a, const point & b) {
if(dcmp(a.x - b.x) == 0)
return a.y < b.y;
else return a.x < b.x;
}
bool operator == (const point &a, const point & b) {
if(dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0)
return true;
return false;
}
Vector operator + (Vector a, Vector b) { return Vector(a.x + b.x, a.y + b.y); }
Vector operator - (Vector a, Vector b) { return Vector(a.x - b.x, a.y - b.y); }
Vector operator * (Vector a, double p) { return Vector(a.x * p, a.y * p); }
Vector operator / (Vector a, double p) { return Vector(a.x / p, a.y / p); }
double dot(Vector a, Vector b) {
return a.x * b.x + a.y * b.y;
}
double len(point a) {
return sqrt(dot(a, a));
}
double sqr(double x) {
return x * x;
}
double dis(point a, point b) {
return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}
// b 在 a 的逆时针为正，夹角为 a 转到 b 的有向角度（sin)
double cross(Vector a, Vector b) {
return a.x * b.y - a.y * b.x;
}
// 求点积再除以模长.
double Angle(Vector a, Vector b) {
return acos(dot(a, b) / len(a) / len(b));
}
// 求法向量.
Vector normal(Vector a) {
double u = len(a);
return Vector(-a.y / u, a.x / u);
}
// 看 tmp 是否在 ab 上
int Onsegment(point tmp, point a, point b) {
if(dcmp(cross(a - tmp, b - tmp)) == 0 && dcmp(dot(a - tmp, b - tmp)) <= 0)
return 1;
return 0;
}
// 前面是线段上，后面是不在线段上相交.
int Line_Intersect(point a, point b, point c, point d) {
double x1 = cross(b - a, c - a), y1 = cross(b - a, d - a);
double x2 = cross(d - c, a - c), y2 = cross(d - c, b - c);
if(!dcmp(x1) || !dcmp(x2) || !dcmp(y1) || !dcmp(y2)) {
return 0;
bool f1 = Onsegment(a, c, d);
bool f2 = Onsegment(b, c, d);
bool f3 = Onsegment(c, a, b);
bool f4 = Onsegment(d, a, b);
// bool f = (f1 | f2 | f3 | f4);
if(f1 && f2) return 1;
if(f3 && f4) return 1;
return 0;
}
if(dcmp(x1) * dcmp(y1) < 0 && dcmp(x2) * dcmp(y2) < 0)
return 1;
return 0;
}
// l1 为直线，l2 为线段, 看是否相交
int seg_inter_line(Line l1, Line l2) {
if(dcmp(cross(l1.e - l2.s, l1.s - l2.s)) * dcmp(cross(l1.e - l2.e, l1.s - l2.e)) <= 0)
return 1;
return 0;
}
// 看直线 l1 和 l2 的位置情况.
int judge_Line(Line l1, Line l2) {
if(dcmp(cross(l1.e - l1.s, l2.e - l2.s)) == 0) {
// 0 : 共线，-1：平行.
if(seg_inter_line(l1, l2) == 1)
return 0;
else
return -1;
}
return 1;
}
// 求直线 l1 与 l2 的交点
point Get_Line(Line l1, Line l2) {
Vector v1 = l1.e - l1.s, v2 = l2.e - l2.s;
double t = cross(l2.s - l1.s, v2) / cross(v1, v2);
return l1.s + v1 * t;
}
#define N 2004
struct Dij {
int hd[N], to[N << 1], nex[N << 1], vis[N], edges;
double val[N << 1], d[N];
struct data {
int x;
double dis;
data(int x = 0, double dis = 0):x(x),dis(dis) {}
bool operator<(const data b) const {
return dis > b.dis;
}
};
priority_queue<data>q;
void add(int u, int v, double c) {
nex[++ edges] = hd[u], hd[u] = edges, to[edges] = v;
val[edges] = c;
}
void dijkstra(int s, int tot) {
for(int i = 0; i <= tot; ++ i)
d[i] = 1000000000, vis[i] = 0;
d[s] = 0;
q.push(data(s, 0));
while(!q.empty()) {
data e = q.top(); q.pop();
if(vis[e.x]) continue;
vis[e.x] = 1;
for(int i = hd[e.x]; i ; i = nex[i]) {
int v = to[i];
if(d[v] > d[e.x] + val[i]) {
d[v] = d[e.x] + val[i];
q.push(data(v, d[v]));
}
}
}
}
void clr(int tot) {
for(int i = 0; i <= tot; ++ i) {
vis[i] = d[i] = 0;
}
for(int i = 0; i <= edges; ++ i) {
nex[i] = to[i] = val[i] = 0;
}
for(int i = 0; i <= tot; ++ i) {
hd[i] = 0;
}
edges = 0;
}
}T;
Line a[N];
point b[N];
int cnt, tot, n;
void add_line(point x, point y) {
a[++ tot] = Line(x, y);
}
int judge(Line tmp) {
// 看 tmp 是否能不碰任何线段.
if(dcmp(dis(tmp.s, tmp.e)) == 0)
return 0;
for(int i = 1; i <= tot ; ++ i) {
if(Line_Intersect(tmp.s, tmp.e, a[i].s, a[i].e) == 1)
return 0;
}
return 1;
}
int main() {
// setIO("input");
// freopen("input.out","w",stdout);
// int fl = 0;
while(1) {
scanf("%d", &n);
if(n == -1) {
break;
}
cnt = tot = 0;
b[++ cnt] = point(0, 5);
b[++ cnt] = point(10, 5);
for(int i = 1; i <= n ; ++ i) {
double x, y[4];
scanf("%lf%lf%lf%lf%lf", &x, &y[0], &y[1], &y[2], &y[3]);
add_line(point(x, 0), point(x, y[0]));
add_line(point(x, y[1]), point(x, y[2]));
add_line(point(x, y[3]), point(x, 10));
b[++ cnt] = point(x, y[0]);
b[++ cnt] = point(x, y[1]);
b[++ cnt] = point(x, y[2]);
b[++ cnt] = point(x, y[3]);
}
// s = 1, t = 2
for(int i = 1; i <= cnt; ++ i) {
for(int j = i + 1; j <= cnt; ++ j) {
if(judge(Line(b[i], b[j]))) {
T.add(i, j, dis(b[i], b[j]));
T.add(j, i, dis(b[i], b[j]));
}
}
}
T.dijkstra(1, cnt);
printf("%.2f\n", T.d[2]);
T.clr(cnt);
}
return 0;
}
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