51NOD1264 线段相交(计算几何)

来源：互联网发布：linux文件夹命名规则编辑：程序博客网时间：2024/06/05 01:01

1264 线段相交基准时间限制：1 秒 空间限制：131072 KB 分值: 0 难度：基础题 收藏  关注给出平面上两条线段的两个端点，判断这两条线段是否相交（有一个公共点或有部分重合认为相交）。 如果相交，输出"Yes"，否则输出"No"。Input第1行：一个数T，表示输入的测试数量(1 <= T <= 1000)第2 - T + 1行：每行8个数，x1,y1,x2,y2,x3,y3,x4,y4。(-10^8 <= xi, yi <= 10^8)(直线1的两个端点为x1,y1 | x2, y2,直线2的两个端点为x3,y3 | x4, y4)Output输出共T行，如果相交输出"Yes"，否则输出"No"。Input示例21 2 2 1 0 0 2 2-1 1 1 1 0 0 1 -1Output示例YesNo

a,b两条线段,a线段(a,b),b线段(c,d),求(ab,ac)叉乘*(ab,ad)叉乘<=0&&(cd,ca)叉乘*(cd,cb)叉乘<=0即线段相交。

#include<iostream>#include<cstring>#include<cstdlib>#include<algorithm>#include<cctype>#include<cmath>#include<ctime>#include<string>#include<stack>#include<deque>#include<queue>#include<list>#include<set>#include<map>#include<cstdio>#include<limits.h>#define MOD 1000000007#define fir first#define sec second#define fin freopen("/home/ostreambaba/文档/input.txt", "r", stdin)#define fout freopen("/home/ostreambaba/文档/output.txt", "w", stdout)#define mes(x, m) memset(x, m, sizeof(x))#define Pii pair<int, int>#define Pll pair<ll, ll>#define INF 1e9+7#define inf 0x3f3f3f3f#define Pi 4.0*atan(1.0)#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1typedef long long ll;typedef unsigned long long ull;const double eps = 1e-9;const int maxn = 1e9;const int maxm = 205;using namespace std;inline int read(){    int x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();    return x*f;}struct point{    double x,y;};double cross(point p0,point p1,point p2){    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);}int main(){    //freopen("/home/ostreambaba/文档/input.txt", "r", stdin);    //freopen("/home/ostreambaba/文档/output.txt", "w", stdout);    int n;    n=read();    point p1,p2,p3,p4;    while(n--){        /*p1.x=read(),p1.y=read(),p2.x=read(),p2.y=read();        p3.x=read(),p3.y=read(),p4.x=read(),p4.y=read();*/        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&p1.x,&p1.y,&p2.x,&p2.y,&p3.x,&p3.y,&p4.x,&p4.y);        double s1,s2,s3,s4;        s1=cross(p3,p4,p1);        s2=cross(p3,p4,p2);        s3=cross(p1,p2,p3);        s4=cross(p1,p2,p4);       // cout<<s1<<" "<<s2<<" "<<s3<<" "<<s4<<endl;        if(s1*s2<=0&&s3*s4<=0){            puts("Yes");        }else{            puts("No");        }    }    return 0;}

0 0