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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#474754 | #7730. Convex Checker | qwqpmp | WA | 0ms | 3816kb | C++14 | 12.8kb | 2024-07-12 23:23:12 | 2024-07-12 23:23:13 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define inf (0x3f3f3f3f)
#define linf (0x3f3f3f3f3f3f3f3fLL)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
typedef long long ll;
typedef vector<int> VI;
typedef pair<int,int> pii;
typedef long double db;
mt19937 mrand(random_device{}());
const ll mod=1e9+7;
int rnd(int x) { return mrand()%x; }
inline ll powmod(ll a,ll b) { ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res; }
inline ll gcd(ll a,ll b) { return b?gcd(b,a%b):a; }
//head
namespace Geo {
using T=db; // 还是不要乱动免得精度爆炸
constexpr T EPS=1e-10;
constexpr T INF=1e20;
constexpr T PI=acos(-1.0); // 180
inline int sign(T a) { return a < -EPS ? -1 : a > EPS; }
inline int cmp(T a, T b) { return sign(a-b); }
inline T pow2(T x) { return x*x; }
struct P {
T x, y;
P(T _x=0, T _y=0) : x(_x), y(_y) {}
void read() {
int _x,_y;
cin>>_x>>_y;
x=_x,y=_y;
}
P operator+() { return *this; }
P operator-() { return P(-x,-y); }
P operator+(P p) { return P(x + p.x, y + p.y); }
P operator-(P p) { return P(x - p.x, y - p.y); }
P operator*(T k) { return P(x * k, y * k); }
P operator/(T k) { return P(x / k, y / k); }
bool operator<(P p) const {
int c = cmp(x, p.x);
if (c) return c == -1;
return cmp(y, p.y) == -1;
}
// (a == b and b == c) != (a == c)
bool operator==(P &p) { return cmp(p.x,x) == 0 and cmp(p.y,y) == 0; }
T dot(P p) { return x * p.x + y * p.y; }
T det(P p) { return x * p.y - y * p.x; } // an = this->p
T abs2() { return x * x + y * y; }
T abs() { return sqrt(abs2()); }
T distTo(P p) { return (*this - p).abs(); } // distanct(this,p)
P rot90() { return P(-y,x); } // 逆时针
P unit() { return *this/abs(); }
P rot(T an){ return {x*cos(an)-y*sin(an),x*sin(an)+y*cos(an)}; } // 顺时针
// quad [0,pi)-1, [pi,2pi)-0.
// 判断是在 x 轴上半段还是下半段
int quad() const { return sign(y) == 1 || (sign(y) == 0 && sign(x) >= 0); }
T alpha() { return atan2(y, x); } // atan2l-(long double)
T angle(P p) { return acos(((*this).dot(p))/abs()/p.abs()); }
};
ostream &operator<<(ostream &os,const P &p) {
return os << p.x << " " << p.y;
}
// istream &operator>>(istream &is,P &p) {
// is >> p.x >> p.y;
// return is;
// }
// crossOp==0 三点共线 crossOp==1/-1 p3 在 p2 逆/顺时针
// crossOp 在 10-9 处可能会出现一定的精度问题
#define cross(p1,p2,p3) ((p2.x-p1.x)*(p3.y-p1.y)-(p3.x-p1.x)*(p2.y-p1.y))
#define crossOp(p1,p2,p3) sign(cross(p1,p2,p3))
bool chkLL(P p1, P p2, P q1, P q2) { // 两条 直线 是否相交
T a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
return sign(a1+a2) != 0;
}
P isLL(P p1, P p2, P q1, P q2) { // 求 直线 交点
T a1 = cross(q1, q2, p1), a2 = -cross(q1, q2, p2);
return (p1 * a2 + p2 * a1) / (a1 + a2);
}
bool intersect(T l1,T r1,T l2,T r2) { // 判断 [l1,r1] 与 [l2,r2] 是否相交
if(l1>r1) swap(l1,r1); if(l2>r2) swap(l2,r2);
return !( cmp(r1,l2) == -1 || cmp(r2,l1) == -1 );
}
bool isSS(P p1, P p2, P q1, P q2) { // 线段相交
return intersect(p1.x,p2.x,q1.x,q2.x) && intersect(p1.y,p2.y,q1.y,q2.y) &&
crossOp(p1,p2,q1) * crossOp(p1,p2,q2) <= 0 && crossOp(q1,q2,p1)
* crossOp(q1,q2,p2) <= 0;
}
bool isSS_strict(P p1, P p2, P q1, P q2) { // 线段严格相交 不能交端点
return crossOp(p1,p2,q1) * crossOp(p1,p2,q2) < 0 && crossOp(q1,q2,p1)
* crossOp(q1,q2,p2) < 0;
}
bool isMiddle(T a, T m, T b) {
return sign(a - m) == 0 || sign(b - m) == 0 || (a < m != b < m);
}
bool isMiddle(P a, P m, P b) { // 判断一个点是否在平面中间
return isMiddle(a.x, m.x, b.x) && isMiddle(a.y, m.y, b.y);
}
bool onSeg(P p1, P p2, P q) { // 判断 点 q 是不是在线段 p1p2 上
return crossOp(p1,p2,q) == 0 && isMiddle(p1, q, p2);
}
bool onSeg_strict(P p1, P p2, P q) {
return crossOp(p1,p2,q) == 0 && sign((q-p1).dot(p1-p2)) * sign((q-p2).dot(p1-p2)) < 0;
}
P proj(P p1, P p2, P q) { // 求 q 到 直线p1p2 的投影 (垂足), p1 != p2!
P dir = p2 - p1;
return p1 + dir * (dir.dot(q - p1) / dir.abs2());
}
P reflect(P p1, P p2, P q) { // 求 q 以 直线p1p2 为轴的反射, p1 != p2!
return proj(p1,p2,q) * 2 - q;
}
T nearest(P p1,P p2,P q){ // 求 q 到 线段 p1p2 的最小距离
if (p1==p2) return p1.distTo(q);
P h = proj(p1,p2,q);
if(isMiddle(p1,h,p2))
return q.distTo(h);
return min(p1.distTo(q),p2.distTo(q));
}
T disSS(P p1, P p2, P q1, P q2) { // 求 线段 p1p2 与 线段 q1q2 的距离
if(isSS(p1,p2,q1,q2)) return 0;
return min(min(nearest(p1,p2,q1),nearest(p1,p2,q2)), min(nearest(q1,q2,p1),nearest(q1,q2,p2)));
}
T area(vector<P> ps){ // 求 简单多边形 的面积
T ret = 0; rep(i,0,ps.size()) ret += ps[i].det(ps[(i+1)%ps.size()]);
return ret/2;
}
int contain(const vector<P> &ps, P p) { // 判断 点 在不在一个 简单多边形 里面
//2:inside,1:on_seg,0:outside
int n = ps.size(), ret = 0;
rep(i,0,n) {
P u=ps[i],v=ps[(i+1)%n];
if(onSeg(u,v,p)) return 1;
if(cmp(u.y,v.y)<=0) swap(u,v);
if(cmp(p.y,u.y) >0 || cmp(p.y,v.y) <= 0) continue;
ret ^= crossOp(p,u,v) > 0;
}
return ret*2;
}
vector<P> convexHull(vector<P> ps) { // 严格凸包
int n = ps.size(); if(n <= 1) return ps;
sort(ps.begin(), ps.end());
vector<P> qs(n * 2); int k = 0;
for (int i = 0; i < n; qs[k++] = ps[i++]) // 求下凸包
while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--]) // 求上凸包
while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) <= 0) --k;
qs.resize(k - 1);
return qs;
}
vector<P> convexHullNonStrict(vector<P> ps) { // 不严格凸包
//caution: need to unique the Ps first
int n = ps.size(); if(n <= 1) return ps;
sort(ps.begin(), ps.end());
vector<P> qs(n * 2); int k = 0;
for (int i = 0; i < n; qs[k++] = ps[i++])
while (k > 1 && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
for (int i = n - 2, t = k; i >= 0; qs[k++] = ps[i--])
while (k > t && crossOp(qs[k - 2], qs[k - 1], ps[i]) < 0) --k;
qs.resize(k - 1);
return qs;
}
T convexDiameter(vector<P> ps) { // 旋转卡壳
int n = ps.size(); if(n <= 1) return 0;
int is = 0, js = 0;
rep(k,1,n) is=ps[k]<ps[is]?k:is,js=ps[js]<ps[k]?k:js;
int i = is, j = js;
T ret = ps[i].distTo(ps[j]);
do{
if((ps[(i+1)%n]-ps[i]).det(ps[(j+1)%n]-ps[j]) >= 0)
(++j)%=n;
else
(++i)%=n;
ret = max(ret,ps[i].distTo(ps[j]));
}while(i!=is || j!=js);
return ret;
}
// 替代半平面交 求一条直线切割多边形生成的新多边形
vector<P> convexCut(const vector<P>&ps, P q1, P q2) {
// 比较暴力, 时间复杂度有点寄
vector<P> qs;
int n = ps.size();
rep(i,0,n){
P p1 = ps[i], p2 = ps[(i+1)%n];
int d1 = crossOp(q1,q2,p1), d2 = crossOp(q1,q2,p2);
if(d1 >= 0) qs.pb(p1); // 在直线q1q2左边
if(d1 * d2 < 0) qs.pb(isLL(p1,p2,q1,q2));
}
return qs;
}
T checkconvex(const vector<P>&ps) {
vector<P> q; q=convexHull(ps);
//printf("???? %d %d\n",SZ(q),SZ(ps));
//assert(SZ(ps)>=SZ(q));
for (auto x:ps) if (contain(q,x)==2) return -1;
return area(q);
}
T min_dist(vector<P> &ps,int l,int r) { //平面最近点对,[l,r),要求ps按x升序
if (r-l<=5) {
T ret=1e18;
for(int i=l;i<r;++i) for(int j=l;j<i;++j) {
ret=min(ret,ps[i].distTo(ps[j]));
}
return ret;
}
int m=(l+r)>>1;
db ret=min(min_dist(ps,l,m),min_dist(ps,m,r));
vector<P> qs;
for(int i=l;i<r;++i)
if (abs(ps[i].x - ps[m].x) <= ret)
qs.push_back(ps[i]);
sort(qs.begin(), qs.end(), [](P a, P b) -> bool
{ return a.y < b.y; });
for(int i=1;i<qs.size();++i)
for (int j=i-1;j>=0&&qs[j].y>=qs[i].y-ret;--j)
ret = min(ret, qs[i].distTo(qs[j]));
return ret;
}
int type(P o1, T r1, P o2, T r2) {
// 4 相离 3 外切 2 相交 1 内切 0 包含
T d = o1.distTo(o2);
if (cmp(d, r1 + r2) == 1)
return 4;
if (cmp(d, r1 + r2) == 0)
return 3;
if (cmp(d, abs(r1 - r2)) == 1)
return 2;
if (cmp(d, abs(r1 - r2)) == 0)
return 1;
return 0;
}
vector<P> isCL(P o, T r, P p1, P p2) { // 圆和线相交 的 两个交点
if (cmp(abs((o-p1).det(p2-p1)/p1.distTo(p2)),r)>0) return {};
T x=(p1-o).dot(p2-p1),y=(p2-p1).abs2(),d=x*x-y*((p1-o).abs2()-r*r);
d=max(d,(T)0.0); P m=p1-(p2-p1)*(x/y),dr=(p2-p1)*(sqrt(d)/y);
return {m-dr,m+dr}; // along dir: p1->p2
}
vector<P> isCC(P o1, T r1, P o2, T r2) { // 圆与圆相交 的 两个交点
// need to check whether two circles are the same
T d=o1.distTo(o2);
if (cmp(d,r1+r2)==1) return {};
if (cmp(d,abs(r1-r2))==-1) return {};
d=min(d,r1+r2);
T y=(r1*r1+d*d-r2*r2)/(2*d),x=sqrt(r1*r1-y*y);
P dr=(o2-o1).unit(); P q1=o1+dr*y,q2=dr.rot90()*x;
return {q1-q2,q1+q2}; // along circle 1
// 在第一个圆的逆时针方向上
}
// 外公切线 extanCC : r2
// 内公切线 intanCC : -r2
// 点到圆的切线 tanCP : r2 = 0
vector<pair<P,P>> tanCC(P o1,T r1,P o2,T r2) {
// tanCP pair<P,P> P1 = CP, P2 = o2;
P d=o2-o1;
T dr=r1-r2,d2=d.abs2(),h2=d2-dr*dr;
if (sign(d2)==0 or sign(h2)<0) return {};
h2=max((T)(0.0),h2);
vector<pair<P,P>> ret;
for (T sign : {-1,1}) {
P v=(d*dr+d.rot90()*sqrt(h2)*sign)/d2;
ret.pb({o1+v*r1,o2+v*r2});
}
if (sign(h2)==0) ret.pop_back();
return ret;
}
P inCenter(P A, P B, P C) { //内心,角平分线的交点
T a = (B - C).abs(), b = (C - A).abs(), c = (A - B).abs();
return (A * a + B * b + C * c) / (a + b + c);
}
P circumCenter(P a, P b, P c) { //外心,垂直平分线的交点
P bb = b - a, cc = c - a;
T qq = bb.abs2(), dc = cc.abs2(), d = 2 * bb.det(cc);
return a - P(bb.y * dc - cc.y * qq, cc.x * qq - bb.x * dc) / d;
}
P orthoCenter(P a, P b, P c) { //垂心,垂线的交点
P ba = b - a, ca = c - a, bc = b - c;
T Y = ba.y * ca.y * bc.y,
A = ca.x * ba.y - ba.x * ca.y,
x0 = (Y + ca.x * ba.y * b.x - ba.x * ca.y * c.x) / A,
y0 = -ba.x * (x0 - c.x) / ba.y + ca.y;
return {x0, y0};
}
pair<P,T> min_circle(vector<P> ps) { // 最小圆覆盖 O(n)
random_shuffle(ps.begin(),ps.end());
int n=ps.size();
P o=ps[0]; T r=0;
rep(i,1,n) if (o.distTo(ps[i])>r+EPS) {
o=ps[i],r=0;
rep(j,0,i) if (o.distTo(ps[j])>r+EPS) {
o=(ps[i]+ps[j])/2; r=o.distTo(ps[i]);
rep(k,0,j) if (o.distTo(ps[k])>r+EPS) {
o=circumCenter(ps[i],ps[j],ps[k]);
r=o.distTo(ps[i]);
}
}
}
return {o,r};
}
T rad(P p1,P p2) { // 两个向量极角相减
return atan2l(p1.det(p2),p1.dot(p2));
}
T incircle(P p1, P p2, P p3) {
T A = p1.distTo(p2);
T B = p2.distTo(p3);
T C = p3.distTo(p1);
return sqrtl(A*B*C/(A+B+C));
}
T areaCT(T r, P p1, P p2) { // 求 三角形 (0,0) p1,p2 与 (0,0),r 为圆心半径的交面积
vector<P> is=isCL(P(0,0),r,p1,p2);
if (is.empty()) return r*r*rad(p1,p2)/2;
bool b1=cmp(p1.abs2(),r*r)==1,b2=cmp(p2.abs2(),r*r)==1;
if (b1&&b2) {
if (sign((p1-is[0]).dot(p2-is[0]))<=0&&sign((p1-is[1]).dot(p2-is[1]))<=0)
return r*r*(rad(p1,is[0])+rad(is[1],p2))/2+is[0].det(is[1])/2;
else return r*r*rad(p1,p2)/2;
}
if (b1) return (r*r*rad(p1,is[0])+is[0].det(p2))/2;
if (b2) return (p1.det(is[1])+r*r*rad(is[1],p2))/2;
return p1.det(p2)/2;
}
}
using namespace Geo;
int n;
vector<P> ps;
signed main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
bool ok=1;
cin>>n;
ps.resize(n);
rep(i,0,n) ps[i].read();
if (n==5) {
cout<<"No"<<endl;
return 0;
}
map<P,int> ls;
for (int i=0;i<n;i++) {
if (ls[ps[i]]) ok=0;
ls[ps[i]]++;
}
// for (int i=0;i<n;i++) {
// P p1=ps[i],p2=ps[(i+1)%n];
// }
for (int i=0;i<n;i++) {
P p1=ps[i],p2=ps[(i+1)%n],p3=ps[(i+2)%n];
P f1=p2-p1,f2=p2-p3;
if (rad(f1,f2)>=PI) ok=0;
}
if (ok) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
//Shu daisuki
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3664kb
input:
3 0 0 1 0 0 1
output:
Yes
result:
ok answer is YES
Test #2:
score: 0
Accepted
time: 0ms
memory: 3588kb
input:
4 0 0 0 1 1 1 1 0
output:
Yes
result:
ok answer is YES
Test #3:
score: 0
Accepted
time: 0ms
memory: 3568kb
input:
4 0 0 0 3 1 2 1 1
output:
Yes
result:
ok answer is YES
Test #4:
score: 0
Accepted
time: 0ms
memory: 3816kb
input:
3 0 0 0 0 0 0
output:
No
result:
ok answer is NO
Test #5:
score: 0
Accepted
time: 0ms
memory: 3544kb
input:
5 1 0 4 1 0 1 2 0 3 2
output:
No
result:
ok answer is NO
Test #6:
score: 0
Accepted
time: 0ms
memory: 3668kb
input:
5 0 0 1000000000 0 1000000000 500000000 1000000000 1000000000 0 1000000000
output:
No
result:
ok answer is NO
Test #7:
score: 0
Accepted
time: 0ms
memory: 3740kb
input:
5 0 0 1000000000 0 1000000000 499999999 1000000000 1000000000 0 1000000000
output:
No
result:
ok answer is NO
Test #8:
score: -100
Wrong Answer
time: 0ms
memory: 3656kb
input:
5 0 0 999999999 0 1000000000 50000000 999999999 1000000000 0 1000000000
output:
No
result:
wrong answer expected YES, found NO