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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#322697 | #7906. Almost Convex | posvii | TL | 1ms | 3832kb | C++14 | 3.0kb | 2024-02-07 15:57:42 | 2024-02-07 15:57:43 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
template<typename G> inline void read(G &x) {x=0;G f=1;char ch=getchar();while((ch<'0'||ch>'9')&&ch!='-') ch=getchar();if(ch=='-') f=-1,ch=getchar();while(ch>='0'&&ch<='9') {x=x*10+(ch^48);ch=getchar();}if(ch=='.') ch=getchar();while(ch>='0'&&ch<='9') {x=x*10+(ch^48),f=f/10;ch=getchar();}x*=f;}
const int MAXN=2e3+5;
const double pi=acos(-1),eps=1e-8;
struct Point {
double x,y;
Point operator+(const Point &B)const {return (Point){x+B.x,y+B.y};}
Point operator-(const Point &B)const {return (Point){x-B.x,y-B.y};}
Point operator*(const double &B)const {return (Point){x*B,y*B};}
Point operator/(const double &B)const {return (Point){x/B,y/B};}
double operator|(const Point &B)const {return x*B.x+y*B.y;}
double operator^(const Point &B)const {return x*B.y-y*B.x;}
bool operator<(const Point &B)const {return fabs(x-B.x)<eps?y+eps<B.y:x+eps<B.x;}
}p[MAXN];
int sign(double x) {return x>eps?1:(x<-eps?-1:0);}
struct Line {
Point a,b;
double slope() {if(a.x==b.x) return 1e9;return (a.y-b.y)/(a.x-b.x);}
friend bool operator<(Line L1,Line L2) {
return (L1.b-L1.a)<(L2.b-L2.a);
}
}L[MAXN];int num;
bool onl(Point A,Line L) {
if(A.x==1e9) return 0;
return (min(L.a.x,L.b.x)<=A.x&&A.x<=max(L.a.x,L.b.x));
}
Point inter(Line a,Line b) {
if(a.slope()==b.slope()) return {1e9,0};
Point res=a.a+(a.b-a.a)*(((b.b-b.a)^(b.a-a.a))/((b.b-b.a)^(a.b-a.a)));
if(a.a.x==a.b.x) {
if(a.a.y<a.b.y) {
if(res.y>a.a.y) return res;
}
else {
if(res.y<a.a.y) return res;
}
}
else {
if(a.a.x<a.b.x) {
if(res.x>a.a.x) return res;
}
else {
if(res.x<a.a.x) return res;
}
}return {1e9,0};
}
double dis(Point A,Point B) {return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}
bool f(Point A,Point B,Point C) {return ((A-B)^(C-B))>eps;}
int n,s[MAXN],tot;
bool vis[MAXN],imp[MAXN];
Line li[MAXN];int cnt;
double ans;
signed main() {
read(n);
for(int i=1;i<=n;++i) read(p[i].x),read(p[i].y);
sort(p+1,p+1+n);
for(int i=1;i<=n;++i) {
while(tot>=2&&!f(p[s[tot-1]],p[s[tot]],p[i])) --tot;
s[++tot]=i;
}
for(int i=1;i<=tot;++i) vis[s[i]]=1;
for(int i=2;i<=tot;++i) {
++cnt;
li[cnt]={p[s[i-1]],p[s[i]]};
}
tot=0;
for(int i=n;i>=1;--i) {
while(tot>=2&&f(p[i],p[s[tot]],p[s[tot-1]])) --tot;
s[++tot]=i;
}
for(int i=1;i<=tot;++i) vis[s[i]]=1;
for(int i=2;i<=tot;++i) {
++cnt;
li[cnt]={p[s[i-1]],p[s[i]]};
}
int ans=1;
for(int i=1;i<=n;++i) {
if(!vis[i]) {
num=0;
for(int j=1;j<=n;++j) {
if(!vis[j]&&j!=i) {
L[++num]={p[i],p[j]};
}
}
sort(L+1,L+num+1);
int now=-1;ans+=cnt;
for(int j=1;j<=cnt;++j) {
imp[j]=0;
if(onl(inter(L[1],li[j]),li[j])) now=j;
}
for(int j=1;j<=num;++j) {
bool flag=0;
for(int k=1;k<=cnt;++k) {
if(onl(inter(L[j],li[k]),li[k])) {
flag=1;
break;
}
}
assert(flag);
while(!onl(inter(L[j],li[now]),li[now])) now=now%cnt+1;
if(!imp[now]) imp[now]=1,--ans;
}
}
}
printf("%d",ans);
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3832kb
input:
7 1 4 4 0 2 3 3 1 3 5 0 0 2 4
output:
9
result:
ok 1 number(s): "9"
Test #2:
score: 0
Accepted
time: 0ms
memory: 3748kb
input:
5 4 0 0 0 2 1 3 3 3 1
output:
5
result:
ok 1 number(s): "5"
Test #3:
score: 0
Accepted
time: 0ms
memory: 3800kb
input:
3 0 0 3 0 0 3
output:
1
result:
ok 1 number(s): "1"
Test #4:
score: 0
Accepted
time: 1ms
memory: 3816kb
input:
6 0 0 3 0 3 2 0 2 1 1 2 1
output:
7
result:
ok 1 number(s): "7"
Test #5:
score: 0
Accepted
time: 0ms
memory: 3828kb
input:
4 0 0 0 3 3 0 3 3
output:
1
result:
ok 1 number(s): "1"
Test #6:
score: -100
Time Limit Exceeded
input:
2000 86166 617851 383354 -277127 844986 386868 -577988 453392 -341125 -386775 -543914 -210860 -429613 606701 -343534 893727 841399 339305 446761 -327040 -218558 -907983 787284 361823 950395 287044 -351577 -843823 -198755 138512 -306560 -483261 -487474 -857400 885637 -240518 -297576 603522 -748283 33...
output:
718