您的位置:首页 > 其它

hdu 3629 Convex

2015-08-11 16:35 369 查看
题意:给你N个点,让你选四个点组成凸多边形,求总的方法数

详细解释:http://blog.sina.com.cn/s/blog_64675f540100ksug.html

#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
const double eps = 1e-12;
const double pi = acos(-1.0);
const double INF = 10001000;
double sqr(double x){
return x*x;
}
int cmp(double x){
if(fabs(x)<eps) return 0;
if(x>0) return 1;
return -1;
}
struct point{
double x,y;
int index;
point(){}
point(double a,double b):x(a),y(b){}
void input(){
scanf("%lf%lf",&x,&y);
}
double angle() {
return atan2(y, x);
}
friend point operator + (const point &a,const point &b){
return point(a.x+b.x,a.y+b.y);
}
friend point operator - (const point &a,const point &b){
return point(a.x-b.x,a.y-b.y);
}
friend bool operator == (const point &a,const point &b){
return (cmp(a.x-b.x)==0)&&(cmp(a.y-b.y))==0;
}
friend point operator * (const point &a,double b){
return point(a.x*b,a.y*b);
}
friend point operator / (const point &a,double b){
return point(a.x/b,a.y/b);
}
friend double operator ^ (const point &a,const point &b)
{
return a.x*b.y - a.y*b.x;
}
double operator *(const point &b)const
{
return x*b.x + y*b.y;
}
double norm(){                              //向量的模长
return sqrt(sqr(x)+sqr(y));
}
};

double det(const point &a,const point &b){      //向量的叉集
return a.x*b.y-a.y*b.x;
}
double dot(const point &a,const point &b){      //向量的点集
return a.x*b.x+a.y*b.y;
}
double dot(const point &a,const point &b,const point &c){      //向量的点集ba 到bc
return dot(a-b,c-b);
}
double dist(const point &a,const point &b){     //两点间的距离
return (a-b).norm();
}
point rotate_point(const point &p,double A){    // 绕原点逆时针旋转 A(弧度)
double tx=p.x,ty=p.y;
return point(tx*cos(A)-ty*sin(A),tx*sin(A)+ty*cos(A));
}
//向量的旋转 //底边线段ab 绕a逆时针旋转角度A,b->b1,sinl是sinA的值。
point rotate_point(double cosl,double sinl,point a, point b){
b.x -= a.x; b.y -= a.y;
point c;
c.x = b.x * cosl - b.y * sinl + a.x;
c.y = b.x * sinl + b.y * cosl + a.y;
return c;
}
double xml(point x,point t1,point t2){    // 如果值为正,(t1-x)在(t2-x)的瞬时间方向
return det((t1-x),(t2-x));
}
double area(point x,point y,point z){
return (det(y-x,z-x));
}
struct line {
point a,b;
line(){}
line(point x,point y):a(x),b(y){}
};
point P_chuizhi_line(point a,point l1,point l2)    // 求一个点,使得ac垂直于l1l2
{
point c;
l2.x -= l1.x; l2.y -= l1.y;
c.x = a.x - l1.x - l2.y + l1.x;
c.y = a.y - l1.y + l2.x + l1.y;
return c;
}
point P_To_seg(point P,line L)                  //点到线段 最近的一个点
{
point result;
double t = ((P-L.a)*(L.b-L.a))/((L.b-L.a)*(L.b-L.a));
if(t >= 0 && t <= 1)
{
result.x = L.a.x + (L.b.x - L.a.x)*t;
result.y = L.a.y + (L.b.y - L.a.y)*t;
}
else
{
if(dist(P,L.a) < dist(P,L.b))
result = L.a;
else result = L.b;
}
return result;
}
double dis_p_to_line(point p,line l){         //点到直线的距离
return fabs(area(p,l.a,l.b))/dist(l.a,l.b);
}
double dis_p_to_seg(point p,line l)            //点到线段的距离
{
return dist(p,P_To_seg(p,l));
}
double dis_pall_seg(point p1, point p2, point p3, point p4)  //平行线段之间的最短距离
{
return min(min(dis_p_to_seg(p1,line(p3,p4)),
dis_p_to_seg(p2, line(p3, p4))),
min(dis_p_to_seg(p3,line(p1, p2)),
dis_p_to_seg(p4,line(p1, p2)))
);
}
bool intbr(line l1,line l2) {            //    线段相交
return
max(l1.a.x,l1.b.x) >= min(l2.a.x,l2.b.x) &&
max(l2.a.x,l2.b.x) >= min(l1.a.x,l1.b.x) &&
max(l1.a.y,l1.b.y) >= min(l2.a.y,l2.b.y) &&
max(l2.a.y,l2.b.y) >= min(l1.a.y,l1.b.y) &&
cmp((l2.a-l1.a)^(l1.b-l1.a))*cmp((l2.b-l1.a)^(l1.b-l1.a)) <= 0 &&
cmp((l1.a-l2.a)^(l2.b-l2.a))*cmp((l1.b-l2.a)^(l2.b-l2.a)) <= 0;
}
point line_inter(point A,point B,point C,point D){ //直线相交交点
point ans;
double a1=A.y-B.y;
double b1=B.x-A.x;
double c1=A.x*B.y-B.x*A.y;

double a2=C.y-D.y;
double b2=D.x-C.x;
double c2=C.x*D.y-D.x*C.y;

ans.x=(b1*c2-b2*c1)/(a1*b2-a2*b1);
ans.y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
return ans;
}

int n;
point ttmp;
point pt1[1001000],pt2[1001000];
bool cmpx(point xx,point yy){
if(cmp(xx.y-yy.y)==0) return xx.x<yy.x;
return xx.y<yy.y;
}
bool cmpd(point xx, point yy){
double db=(xx-ttmp)^(yy-ttmp);
if(cmp(db)==0) return dist(xx,ttmp)<dist(yy,ttmp);
if(cmp(db)>0) return 1;
else return 0;
}
point grp1[1001000],grp2[1001000];
int Graham(point* grp,point *pt,int n){             //凸包
int top=1;
sort(pt,pt+n,cmpx);ttmp=pt[0];
sort(pt+1,pt+n,cmpd);
grp[0]=pt[0];
grp[1]=pt[1];
for(int i=2;i<n;i++){
while(top>0){
double db=(pt[i]-grp[top])^(grp[top]-grp[top-1]);
if(cmp(db)>=0) top--;
else break;
}
grp[++top]=pt[i];
}
return top+1;
}
double rotating_calipers(point* grp ,int len){ //旋转卡壳求凸包直径
int i=0,j=1;
double ans=0;
while(i<len){
while(area(grp[i],grp[i+1],grp[(j+1)%len])>
area(grp[i],grp[i+1],grp[j])) j=(j+1)%len;
ans=max(ans,max(dist(grp[i],grp[j]),dist(grp[i+1],grp[j])));
i++;
}
return ans;
}
double rotating_calipers2(point* grp1,int len1,point* grp2,int len2){             //旋转卡壳 求两个凸包的最远距离
int p=0,q=0;
for(int i=0;i<len1;i++) if(grp1[i].y<grp1[p].y) p=i;
for(int i=0;i<len2;i++) if(grp2[i].y>grp2[q].y) q=i;
double ans=1e99,tmp;
grp1[len1]=grp1[0];//避免取模
grp2[len2]=grp2[0];//避免取模
for(int i=0;i<len1;i++){
while(tmp=cmp(area(grp1[p],grp1[p+1],grp2[q+1])-
area(grp1[p],grp1[p+1],grp2[q]))>0)   q=(q+1)%len2;
if(tmp==0) ans=min(ans,dis_pall_seg(grp1[p],grp1[p+1],grp2[q],grp2[q+1]));
else ans=min(ans,dis_p_to_seg(grp2[q],line(grp1[p],grp1[p+1])));
p=(p+1)%len1;
}

return ans;
}
double rotating_calipers3(point* grp ,int len){             //旋转卡壳求凸包宽度
int p=0,q=0;
int tmp;
for(int i=0;i<len;i++) if(grp[i].y<grp[p].y) p=i;
for(int j=0;j<len;j++) if(grp[j].y>grp[q].y) q=j;
double ans=1e30;
for(int i=0;i<len;i++){
while(tmp=cmp(area(grp[p],grp[p+1],grp[(q+1)%len])-
area(grp[p],grp[p+1],grp[q]))>0) q=(q+1)%len;
ans=min(ans,dis_p_to_line(grp[q],line(grp[p],grp[(p+1)%len])));
p=(p+1)%len;
}
return ans;
}
double rotating_calipers4(point* grp,int len){
double ans;
int xmin=0,xmax=0,ymin=0,ymax=0;
for(int i=0;i<len;i++) if(cmp(grp[xmin].x-grp[i].x)>0) xmin=i;
for(int i=0;i<len;i++) if(cmp(grp[xmax].x-grp[i].x)<0) xmax=i;
for(int i=0;i<len;i++) if(cmp(grp[ymin].y-grp[i].y)>0) ymin=i;
for(int i=0;i<len;i++) if(cmp(grp[ymax].y-grp[i].y)<0) ymax=i;
ans=(grp[ymax].y-grp[ymin].y)*(grp[xmax].x-grp[xmin].x);
grp[len]=grp[0];
for(int i=0;i<len;i++){
while(cmp(area(grp[ymin],grp[ymin+1],grp[ymax+1])-
area(grp[ymin],grp[ymin+1],grp[ymax]))>=0) ymax=(ymax+1)%len;
while(cmp(dot(grp[xmax+1],grp[ymin],grp[ymin+1])-
dot(grp[xmax],grp[ymin],grp[ymin+1]))>=0) xmax=(xmax+1)%len;
if(i==0) xmin=xmax;
while(cmp(dot(grp[xmin+1],grp[ymin+1],grp[ymin])-
dot(grp[xmin],grp[ymin+1],grp[ymin]))>=0) xmin=(xmin+1)%len;
double L1=dis_p_to_line(grp[ymax],line(grp[ymin],grp[ymin+1]));
point a=P_chuizhi_line(grp[xmin],grp[ymin],grp[ymin+1]);
double L2=dis_p_to_line(grp[xmax],line(grp[xmin],a));
if(ans>L1*L2){
ans=L1*L2;

}
ymin=(ymin+1)%len;
}
return ans;
}
struct node{
double angle;
}fuck[1000];
bool nodecmp(node a,node b){
return a.angle<b.angle;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif // ONLINE_JUDGE
int t;cin>>t;
while(t--){
scanf("%d",&n);
for(int i=0;i<n;i++) pt1[i].input();
long long ans=1ll*n*(n-1)*(n-2)*(n-3)/24;
for(int i=0;i<n;i++){
int cnt=0;
for(int j=0;j<n;j++) if(i!=j){
fuck[cnt++].angle=(pt1[j]-pt1[i]).angle()+pi;
}
sort(fuck,fuck+cnt,nodecmp);
for(int j=cnt;j<2*cnt;j++) fuck[j].angle=fuck[j-cnt].angle+2*pi;
int st=1;
long long tmp=0;
for(int j=0;j<cnt;j++){
while(fuck[st].angle-fuck[j].angle<pi) st++;
if(st-j-1<2) continue;
tmp+=(st-j-1)*(st-j-2)/2;
}
ans-=(1ll*(n-1)*(n-2)*(n-3)/6-tmp);
}
printf("%lld\n",ans);
}
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: