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有向面积裸题

 

所有有向三角形面积加起来就行

Code:

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<queue>
  6 #include<iostream>
  7 #define ms(a,b) memset(a,b,sizeof a)
  8 #define rep(i,a,n) for(int i = a;i <= n;i++)
  9 #define per(i,n,a) for(int i = n;i >= a;i--)
 10 #define inf 2147483647
 11 using namespace std;
 12 typedef long long ll;
 13 typedef double D;
 14 #define eps 1e-8
 15 ll read() {
 16     ll as = 0,fu = 1;
 17     char c = getchar();
 18     while(c < '0' || c > '9') {
 19         if(c == '-') fu = -1;
 20         c = getchar();
 21     }
 22     while(c >= '0' && c <= '9') {
 23         as = as * 10 + c - '0';
 24         c = getchar();
 25     }
 26     return as * fu;
 27 }
 28 //head
 29 #define P point
 30 struct point {
 31     D x,y;
 32     point(){}
 33     point(D X,D Y):x(X),y(Y){}
 34     D len() {return sqrt(x*x+y*y);}
 35     D tan() {return y/x;}
 36     D sin() {return len() / y;}
 37     D cos() {return len() / x;}
 38     void print() {printf("%.2lf %.2lf\n",x,y);}
 39     friend inline point operator + (const point &a,const point &b) {
 40         return point(a.x+b.x,a.y+b.y);
 41     }
 42     friend inline point operator - (const point &a,const point &b) {
 43         return point(a.x-b.x,a.y-b.y);
 44     }
 45     //放缩
 46     friend inline point operator * (const point &a,const D &b) {
 47         return point(a.x*b,a.y*b);
 48     }
 49     //叉积
 50     friend inline D operator * (const point &a,const point &b) {
 51         return a.x*b.y-a.y*b.x;
 52     }
 53     //点积
 54     friend inline D operator / (const point &a,const point &b) {
 55         return a.x*b.x+a.y*b.y;
 56     }
 57 };
 58 struct line {
 59     D k,b;
 60     void init(point x,point y) {
 61         k = (y.y-x.y)/(y.x-x.x);
 62         b = x.y - k * x.x;
 63     }
 64     D YY(D X) {return k*X+b;}
 65     D XX(D Y) {return (Y-b)/k;}
 66 };
 67 struct yuan {
 68     D r,x,y;
 69     yuan(){}
 70     yuan(int R,int X,int Y):r(R),x(X),y(Y){}
 71 };
 72 
 73 bool ONSEG(point a,point b,point p) {
 74     return ((a-b).len() == (a-p).len() + (p-b).len());
 75 }
 76 D TRIAREA(point a,point b,point c) {
 77     return ((a-b)*(a-c)) / 2.0;
 78 }
 79 #define ONLINE(a,b,c) (((a-b)*(a-c)) == 0)
 80 
 81 #define sign(x) (x) > 0 ? 1 : ((x) < 0 ? -1 : 0)
 82 // 1  0 -1
 83 // 锐 直 钝
 84 int ANGDIR(point a,point b,point p) {
 85     D ans = (p-a)*(p-b);
 86     return sign(ans);
 87 }
 88 
 89 D dis(point a,point b,point p) {
 90     if(ANGDIR(b,p,a) == -1) return (p-a).len();
 91     if(ANGDIR(a,p,b) == -1) return (p-b).len();
 92     return ((p-a)*(p-b)) / (a-b).len();
 93 }
 94 D dis(point a,line l) {
 95     return (l.k * a.x - a.y + l.b) / sqrt(l.k*l.k+1);
 96 }
 97 
 98 int cross(P a,P b,P c,P d) {
 99     if(ONLINE(a,b,c) ^ ONLINE(a,b,d)) return 1;
100     if(ONLINE(c,d,a) ^ ONLINE(c,d,b)) return 1;
101     if(ONLINE(a,b,c) & ONLINE(a,b,d)) return -1;
102     if(ONLINE(c,d,a) & ONLINE(c,d,b)) return -1;
103     D J1 = ((c-d)*(c-a)) * ((c-d)*(c-b));
104     D J2 = ((a-b)*(a-c)) * ((a-b)*(a-d));
105     if(J1 < 0 && J2 < 0) return 1;
106     return 0;
107 }
108 
109 point Cross(point a,point b,point c,point d) {
110     if(ONLINE(a,b,c)) return c;
111     if(ONLINE(a,b,d)) return d;
112     if(ONLINE(c,d,a)) return a;
113     if(ONLINE(c,d,b)) return b;
114     D S1 = (a-c)*(a-d),S2 = (b-d) * (b-c);
115     point tmp = (b-a) * (S1 / (S1+S2));
116     return a + tmp;
117 }
118 //CP
119 const int N = 10005;
120 point p[N];
121 int main() {
122     while(1) {
123         int n = read();
124         if(!n) return 0;
125         rep(i,1,n) scanf("%lf %lf",&p[i].x,&p[i].y);
126         D res = 0.0;
127         rep(i,1,n) (i ^ n) ? res += p[i] * p[i+1] : res += p[n] * p[1];
128         printf("%.0lf\n",fabs(res) / 2.0);
129     }
130 }

 

posted @ 2018-11-21 17:27  白怀潇  阅读(150)  评论(0)  编辑  收藏  举报