poj3304 Segments

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多组数据:给n条线段 判断是否有一条直线能与所有的线段产生交♂点

首先n=1/n=2显然

然后考虑如果有一种方案是从中间穿过去 那么一定有某种方案

使得这条直线绕某个点旋转某个角度

然后交上一个线段的端点

所以枚举端点对然后去交线段就行

注意有一个交不上就验证下一个 全都不成立再输出no等等

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 read() {scanf("%lf%lf",&x,&y);}
 39     void print() {printf("%.2lf %.2lf\n",x,y);}
 40     friend inline point operator + (const point &a,const point &b) {
 41         return point(a.x+b.x,a.y+b.y);
 42     }
 43     friend inline point operator - (const point &a,const point &b) {
 44         return point(a.x-b.x,a.y-b.y);
 45     }
 46     //放缩
 47     friend inline point operator * (const point &a,const D &b) {
 48         return point(a.x*b,a.y*b);
 49     }
 50     //叉积
 51     friend inline D operator * (const point &a,const point &b) {
 52         return a.x*b.y-a.y*b.x;
 53     }
 54     //点积
 55     friend inline D operator / (const point &a,const point &b) {
 56         return a.x*b.x+a.y*b.y;
 57     }
 58     bool operator == (const point &o) const {
 59         return (x == o.x) && (y == o.y);
 60     }
 61 };
 62 struct yuan {
 63     D r,x,y;
 64     yuan(){}
 65     yuan(int R,int X,int Y):r(R),x(X),y(Y){}
 66 };
 67 struct line {
 68     point x,y;
 69 };
 70 bool ONSEG(point a,point b,point p) {
 71     return ((a-b).len() == (a-p).len() + (p-b).len());
 72 }
 73 D TRIAREA(point a,point b,point c) {
 74     return ((a-b)*(a-c)) / 2.0;
 75 }
 76 #define ONLINE(a,b,c) (((a-b)*(a-c)) == 0)
 77 #define sign(x) (x) > 0 ? 1 : ((x) < 0 ? -1 : 0)
 78 // 1  0 -1
 79 // 锐 直 钝
 80 int ANGDIR(point a,point b,point p) {
 81     D ans = (p-a)*(p-b);
 82     return sign(ans);
 83 }
 84 D dis(point a,point b,point p) {
 85     if(ANGDIR(b,p,a) == -1) return (p-a).len();
 86     if(ANGDIR(a,p,b) == -1) return (p-b).len();
 87     return ((p-a)*(p-b)) / (a-b).len();
 88 }
 89 int cross(P a,P b,P c,P d) {
 90     if(ONLINE(a,b,c) ^ ONLINE(a,b,d)) return 1;
 91     if(ONLINE(c,d,a) ^ ONLINE(c,d,b)) return 1;
 92     if(ONLINE(a,b,c) & ONLINE(a,b,d)) return -1;
 93     if(ONLINE(c,d,a) & ONLINE(c,d,b)) return -1;
 94     D J1 = ((c-d)*(c-a)) * ((c-d)*(c-b));
 95     D J2 = ((a-b)*(a-c)) * ((a-b)*(a-d));
 96     if(J1 < 0 && J2 < 0) return 1;
 97     return 0;
 98 }
 99 point Cross(point a,point b,point c,point d) {
100     if(ONLINE(a,b,c)) return c;
101     if(ONLINE(a,b,d)) return d;
102     if(ONLINE(c,d,a)) return a;
103     if(ONLINE(c,d,b)) return b;
104     D S1 = (a-c)*(a-d),S2 = (b-d) * (b-c);
105     point tmp = (b-a) * (S1 / (S1+S2));
106     return a + tmp;
107 }
108 //CP
109 int n;
110 const int N = 2006;
111 point p[N<<1];
112 
113 bool check(int x,int y) {
114     point a,b;
115     rep(k,1,n) {
116         a = p[k-1<<1|1],b = p[k<<1];
117         if(((a-p[x])*(a-p[y])) * ((b-p[x])*(b-p[y])) > 0.0) return 0;
118     }
119     return 1;
120 }
121 
122 bool solve() {
123     n = read();
124     rep(i,1,n<<1) scanf("%lf%lf",&p[i].x,&p[i].y);
125     rep(i,1,n<<1) rep(j,i+1,n<<1) {
126         if(p[i] == p[j]) continue;
127         if(check(i,j)) return 1;
128     }
129     return 0;
130 }
131 
132 int main() {
133     int T = read();
134     while(T--) solve() ? puts("Yes!") : puts("No!");
135     return 0;
136 }

依然一大堆函数没用qwq

 

posted @ 2018-11-21 17:20  白怀潇  阅读(133)  评论(0)  编辑  收藏  举报