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  • How I Mathematician Wonder What You Are! POJ

    How I Mathematician Wonder What You Are!

     POJ - 3130

    题意:判断多边形是否有内核

    思路:半平面交题,逆时针存入

     1 // 
     2 // Created by HJYL on 2020/2/6.
     3 //
     4 #include<iostream>
     5 #include<stdio.h>
     6 #include<algorithm>
     7 #include<math.h>
     8 using namespace std;
     9 typedef long long ll;
    10 typedef double db;
    11 const int N=207;
    12 const db eps=1e-7;
    13 int sign(db k){if(k>eps) return 1;if(k<-eps) return -1; return 0;}
    14 int dcmp(db k1,db k2){return sign(k1-k2);}
    15 struct point{
    16     db x,y;
    17     point operator - (const point &k)const{
    18         return (point){x-k.x,y-k.y};
    19     }
    20     point operator + (const point &k)const{
    21         return (point){x+k.x,y+k.y};
    22     }
    23     point operator * (const db &k)const{
    24         return (point){x*k,y*k};
    25     }
    26     point operator / (const db &k)const{
    27         return (point){x/k,y/k};
    28     }
    29     db angle(){return atan2(y,x);}
    30     void print(){printf("(%f,%f)
    ",x,y);}
    31 }P[N];
    32 db cross(point k1,point k2){
    33     return k1.x*k2.y-k1.y*k2.x;
    34 }
    35 db dot(point k1,point k2){
    36     return k1.x*k2.x+k1.y*k2.y;
    37 }
    38 struct line{
    39     point s,e;
    40     db angle(){return (e-s).angle();}
    41 }L[N],dq[N];
    42 point getLL(line k1,line k2){
    43     db w1=cross(k1.s-k2.s,k2.e-k2.s),w2=cross(k2.e-k2.s,k1.e-k2.s);
    44     return (k1.s*w2+k1.e*w1)/(w1+w2);
    45 }
    46 bool onRight(line k,point p){
    47     return sign((cross(p-k.s,k.e-k.s)))>0;
    48 }
    49 bool cmp(line k1,line k2){
    50     if(dcmp(k1.angle(),k2.angle())==0) return onRight(k2,k1.s);
    51     return k1.angle()<k2.angle();
    52 }
    53 int tot;
    54 bool halfplaneinsert(){
    55     sort(L+1,L+1+tot,cmp);
    56     int cnt=0;
    57     for(int i=1;i<=tot;i++){
    58         if(i<tot&&dcmp(L[i].angle(),L[i+1].angle())==0) continue;
    59         L[++cnt]=L[i];
    60     }
    61     int tail=-1,head=0;
    62     for(int i=1;i<=cnt;i++){
    63         while(tail-head>=1&&onRight(L[i],getLL(dq[tail],dq[tail-1]))) tail--;
    64         while(tail-head>=1&&onRight(L[i],getLL(dq[head],dq[head+1]))) head++;
    65         dq[++tail]=L[i];
    66     }
    67     while(tail-head>=1&&onRight(dq[head],getLL(dq[tail],dq[tail-1]))) tail--;
    68     while(tail-head>=1&&onRight(dq[tail],getLL(dq[head],dq[head+1]))) head++;
    69     return tail-head>=2;
    70 }
    71 int main()
    72 {
    73     int n;
    74     while(~scanf("%d",&n)&&n) {
    75             for (int i = 1; i <= n; i++) {
    76                 scanf("%lf%lf", &P[i].x, &P[i].y);
    77             }
    78             tot = 0;
    79             for (int i = 1; i < n; i++) {
    80                 L[++tot] = (line) {P[i], P[i + 1]};
    81             }
    82             L[++tot] = (line) {P[n], P[1]};
    83             if (halfplaneinsert()) printf("1
    ");
    84             else printf("0
    ");
    85         }
    86 }
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  • 原文地址:https://www.cnblogs.com/Vampire6/p/12274698.html
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