POJ 2653 线段相交问题

来源：互联网发布：矩阵化为行最简形编辑：程序博客网时间：2024/04/29 22:40

题意：按顺序给出一堆线段的端点坐标求最后没被覆盖的线段的序号被覆盖（即后放入与先放入的相交）利用叉积判断即可

#include <iostream>#include "stdio.h"#include "stdlib.h"#include "string.h"#include "math.h"#include "algorithm"#include <queue>using namespace std;#define modulo 1000000007const double eps=1e-8;int t,pos,first,last,vis[100001],ans,answer[1001];struct ed {    double x1,y1,x2,y2;}edge[100001];double mult(double x1,double y1,double x2,double y2,double x3,double y3){    return (x1-x3)*(y2-y3)-(y1-y3)*(x2-x3);}int  sgn(double x){    if(fabs(x)<eps) return 0;    if(x<0) return -1;    else return 1;}bool isinter(ed l1,ed l2){    return        max(l1.x1,l1.x2)>=min(l2.x1,l2.x2)&&        max(l2.x1,l2.x2)>=min(l1.x1,l1.x2)&&        max(l1.y1,l1.y2)>=min(l2.y1,l2.y2)&&        max(l2.y1,l2.y2)>=min(l1.y1,l1.y2)&&        sgn(mult(l1.x1,l1.y1,l1.x2,l1.y2,l2.x1,l2.y1))*sgn(mult(l1.x1,l1.y1,l1.x2,l1.y2,l2.x2,l2.y2))<=0&&        sgn(mult(l2.x1,l2.y1,l2.x2,l2.y2,l1.x1,l1.y1))*sgn(mult(l2.x1,l2.y1,l2.x2,l2.y2,l1.x2,l1.y2))<=0;}int main(void){    while(scanf("%d",&t)&&t)    {        pos=0;        ans=0;        memset(vis,0,sizeof vis);        for(int i=0;i<t;i++)            scanf("%lf%lf%lf%lf",&edge[i].x1,&edge[i].y1,&edge[i].x2,&edge[i].y2);        for(int i=0;i<t;i++)            for(int j=i+1;j<t;j++)                if(isinter(edge[i],edge[j])) {                    vis[i]=1;                    break;                }        printf("Top sticks:");        for(int i=0;i<t;i++)            if(!vis[i]) answer[pos++]=i+1;        for(int i=0;i<pos-1;i++)            printf(" %d,",answer[i]);        printf(" %d.\n",answer[pos-1]);    }}

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