Find them, Catch them POJ

题意：给一些木棍（两个端点）判断两个木棍是否连接（间接连接也可以，间接连结就是并查集好用）

分析：判断线段相交的模板，和并查集基本应用。

收获：模板+1；

并查集的rank+1；

从卢神的代码中学到了处理并查集find函数的正确操作

以前都是

int find_father(int x){    return x==father[x]?x:find_father(father[x]);}

很简短，觉得自己很帅，其实不存在的。这要会浪费很多时间。

正确的应该是：

int find_father(int x){    if(father[x]==x) return x;    return father[x]=find_father(father[x]);}

这样在向上求父亲的时候会顺路把经过的点father[x]赋值

#include <iostream>#include <algorithm>#include <cstdio>#include <string>using namespace std;int father[200];void Init( ){    for(int i=0; i<=199; i++)    {        father[i]=i;    }    return;}int find_father(int x){    if(father[x]==x) return x;    return father[x]=find_father(father[x]);}void Union(int a,int b){    a=find_father(a);    b=find_father(b);    father[b]=a;}//mo  bannnconst double eps=1e-10;struct point{    double x, y;} sb[20],se[20];double min(double a, double b){    return a < b ? a : b;}double max(double a, double b){    return a > b ? a : b;}bool inter(point a, point b, point c, point d){    if (    min(a.x, b.x) > max(c.x, d.x) ||            min(a.y, b.y) > max(c.y, d.y) ||            min(c.x, d.x) > max(a.x, b.x) ||            min(c.y, d.y) > max(a.y, b.y) ) return 0;    double h, i, j, k;    h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);    i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);    j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);    k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);    return h * i <= eps && j * k <= eps;}//mo bannnint main (){    int m;    while(scanf("%d",&m)&&m)    {        for(int i=1; i<=m; i++)        {            scanf("%lf%lf%lf%lf",&sb[i].x,&sb[i].y,&se[i].x,&se[i].y);        }        Init();        for(int i=1; i<=m; i++)        {            for(int j=i+1; j<=m; j++)            {                if(inter(sb[i],se[i],sb[j],se[j]))                {                    Union(i,j);                }            }        }        int a,b;        while(scanf("%d%d",&a,&b)&&a&&b)        {            a=find_father(a);            b=find_father(b);            if(a==b) cout << "CONNECTED\n";            else cout << "NOT CONNECTED\n";        }    }    return 0;}




                                             
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