hdu 1588 线段相交+并查集

判断第k个线段的集合中一共有几条线段。

先用并查集将相交的线段合并记录，最后查询sum数组即可。

#include<iostream>#include<string>#include<cstring>using namespace std;struct node{double x;double y;}s[2004];struct edge{node a,b;int num;}E[2005];int fa[2005];int sum[2005]; int T,n,cha;char c;int m=0;double direction (node p0, node p1, node p2) {      return ((p2.x - p0.x)*(p1.y - p0.y) - (p1.x - p0.x)*(p2.y - p0.y));  }    bool on_segment (node p0, node p1, node p2)//判断线段相交{      double minx, maxx, miny, maxy;      minx = min(p0.x, p1.x);      maxx = max(p0.x, p1.x);      miny = min(p0.y, p1.y);      maxy = max(p0.y, p1.y);      if (p2.x >= minx && p2.x <= maxx && p2.y >= miny && p2.y <= maxy)          return true;      else          return false;  }  bool segments_intersect (node p1, node p2, node p3, node p4) {      double d1, d2, d3, d4;      d1 = direction(p3, p4, p1);      d2 = direction(p3, p4, p2);      d3 = direction(p1, p2, p3);      d4 = direction(p1, p2, p4);      if (((d1 < 0 && d2 > 0) || (d1 > 0 && d2 < 0)) && ((d3 < 0 && d4 > 0) || (d3 > 0 && d4 < 0)))          return true;      else if (d1 == 0 && on_segment(p3, p4, p1))          return true;      else if (d2 == 0 && on_segment(p3, p4, p2))          return true;      else if (d3 == 0 && on_segment(p1, p2, p3))          return true;      else if (d4 == 0 && on_segment(p1, p2, p4))          return true;      else          return false;  } //并查集操作 void init(){for(int i=1;i<=n;i++){   fa[i]=i;   sum[i]=1;}}int find(int x){if(x!=fa[x])fa[x]=find(fa[x]);return fa[x];}void combine(int x,int y){int xx=find(x);int yy=find(y);if(xx==yy)return ;    fa[xx]=yy;    sum[yy]+=sum[xx];}int main(){cin>>T;while(T--){      cout<<(m++?"\n":"");cin>>n;init();int cixu=1;for(int i=1;i<=n;i++){cin>>c;if(c=='P'){cin>>E[cixu].a.x>>E[cixu].a.y>>E[cixu].b.x>>E[cixu].b.y;E[cixu].num=cixu;for(int j=1;j<cixu;j++){if(segments_intersect(E[j].a,E[j].b,E[cixu].a,E[cixu].b)){combine(E[j].num,E[cixu].num);}}cixu++;}    if(c=='Q'){cin>>cha;cout<<sum[find(cha)]<<endl;}} }return 0;}