题目链接:https://vjudge.net/problem/POJ-2653
题目:
思路:暴力判断两线段间是否相交,若相交则去掉前面的线段,将对应棍子的编号输出即可
1 // 2 // Created by HJYL on 2020/1/13. 3 // 4 #include<iostream> 5 #include<cstdio> 6 #include<cstring> 7 #include<queue> 8 #include<cmath> 9 #define eps 1e-8 10 #define Inf 0x7fffffff 11 //#include<bits/stdc++.h> 12 using namespace std; 13 const int maxn=1e5+100; 14 struct Point{ 15 double x,y; 16 }; 17 double min(double a, double b) { return a < b ? a : b; } 18 19 double max(double a, double b) { return a > b ? a : b; } 20 double Cross(Point &sp, Point &ep, Point &op) 21 { 22 return (sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y); 23 } 24 bool IsSegmentIntersect(Point a, Point b, Point c, Point d) 25 { 26 if( min(a.x, b.x) > max(c.x, d.x) || 27 min(a.y, b.y) > max(c.y, d.y) || 28 min(c.x, d.x) > max(a.x, b.x) || 29 min(c.y, d.y) > max(a.y, b.y) ) 30 return 0; 31 32 double h, i, j, k; 33 34 h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x); 35 i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x); 36 j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x); 37 k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x); 38 39 return h * i <= eps && j * k <= eps; 40 } 41 42 int main() 43 { 44 int T; 45 bool flag[maxn]; 46 while(~scanf("%d",&T)) { 47 memset(flag,0, sizeof(flag)); 48 if(T==0) 49 break; 50 Point p[maxn],b[maxn]; 51 for(int i=0;i<T;i++){ 52 scanf("%lf%lf%lf%lf", &p[i].x, &p[i].y, &b[i].x, &b[i].y); 53 } 54 //for(int i=0;i<=pos;i++) 55 // cout<<p[i].x<<" "<<p[i].y<<endl; 56 for(int i=0;i<T;i++) 57 { 58 for(int j=i+1;j<T;j++) 59 { 60 61 if(IsSegmentIntersect(p[i],b[i],p[j],b[j])) 62 { 63 flag[i]=1; 64 break; 65 } 66 } 67 } 68 int gg[maxn]={0}; 69 int re=0; 70 for(int i=0;i<T;i++) 71 if(!flag[i]) gg[re++]=i+1; 72 printf("Top sticks:"); 73 for(int i=0;i<re-1;i++) 74 printf(" %d,",gg[i]); 75 printf(" %d.\n",gg[re-1]); 76 } 77 return 0; 78 }
原文:https://www.cnblogs.com/Vampire6/p/12193106.html