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NYOJ 227 && poj1556 判断直线相交

2017-04-28 19:47 351 查看
总共4*n+1个点个点。
根据线段交去判断会不会冲突,可以直接连的就建立一条路径。

然后求最短路。
因为点很小,所以无论哪种最短路算法都可使用。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>

using namespace std;

const double eps = 1e-8;
int sgn(double x)
{
if(fabs(x) < eps)return 0;
if(x < 0) return -1;
else return 1;
}
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y)
{
x = _x;y = _y;
}
Point operator -(const Point &b)const
{
return Point(x - b.x,y - b.y);
}
double operator ^(const Point &b)const
{
return x*b.y - y*b.x;
}
double operator *(const Point &b)const
{
return x*b.x + y*b.y;
}
};
struct Line
{
Point s,e;
Line(){}
Line(Point _s,Point _e)
{
s = _s;e = _e;
}
};
//判断线段相交
bool inter(Line l1,Line l2)
{
return
max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s)) <= 0 &&
sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s)) <= 0;
}
double dist(Point a,Point b)
{
return sqrt((b-a)*(b-a));
}
const int MAXN = 100;
Line line[MAXN];
double dis[MAXN][MAXN];
const double INF = 1e20;
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
double x,y1,y2,y3,y4;
while(scanf("%d",&n) == 1)
{
if(n == -1) break;
for(int i = 1;i <= n;i++)
{
scanf("%lf%lf%lf%lf%lf",&x,&y1,&y2,&y3,&y4);
line[2*i-1] = Line(Point(x,y1),Point(x,y2));
line[2*i] = Line(Point(x,y3),Point(x,y4));
}
for(int i = 0;i <= 4*n+1;i++)
for(int j = 0;j <= 4*n+1;j++)
{
if(i == j)dis[i][j] = 0;
else dis[i][j] = INF;
}
for(int i = 1;i <= 4*n;i++)
{
int lid = (i+3)/4;
bool flag = true;
Point tmp;
if(i&1)tmp = line[(i+1)/2].s;
else tmp = line[(i+1)/2].e;
for(int j = 1;j < lid;j++)
if(inter(line[2*j-1],Line(Point(0,5),tmp)) == false
&& inter(line[2*j],Line(Point(0,5),tmp)) == false)
flag = false;
if(flag)dis[0][i] =dis[i][0] = dist(Point(0,5),tmp);
flag = true;
for(int j = lid+1;j <= n;j++)
if(inter(line[2*j-1],Line(Point(10,5),tmp)) == false
&& inter(line[2*j],Line(Point(10,5),tmp)) == false)
flag = false;
if(flag)dis[i][4*n+1] =dis[4*n+1][i] = dist(Point(10,5),tmp);
}
for(int i = 1;i <= 4*n;i++)
for(int j = i+1;j <=4*n;j++)
{
int lid1 = (i+3)/4;
int lid2 = (j+3)/4;
bool flag = true;
Point p1,p2;
if(i&1)p1 = line[(i+1)/2].s;
else p1 = line[(i+1)/2].e;
if(j&1)p2 = line[(j+1)/2].s;
else p2 = line[(j+1)/2].e;
for(int k = lid1+1;k < lid2;k++)
if(inter(line[2*k-1],Line(p1,p2)) == false
&& inter(line[2*k],Line(p1,p2)) == false)
flag = false;
if(flag) dis[i][j] = dis[j][i] = dist(p1,p2);
}
bool flag = true;
for(int i = 1;i <= n;i++)
if(inter(line[2*i-1],Line(Point(0,5),Point(10,5))) == false
&& inter(line[2*i],Line(Point(0,5),Point(10,5))) == false)
flag = false;
if(flag)dis[0][4*n+1] = dis[4*n+1][0] = 10;
for(int k = 0;k <= 4*n+1;k++)
for(int i = 0;i <= 4*n+1;i++)
for(int j = 0;j <= 4*n+1;j++)
if(dis[i][k] + dis[k][j] < dis[i][j])
dis[i][j] = dis[i][k] + dis[k][j];
printf("%.2lf\n",dis[0][4*n+1]);
}

return 0;
}
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