Treasure Hunt--POJ 1066

1、题目类型：计算几何、线段相交。

2、解题思路：分析，读题后第一想法是从treasure点多边形向外做BFS直到到达边界，但区域内的各个不规则多边形无法确定；后发现只要在区域的四周的焦点间中点做与treasure点的线段，获取其最小焦点就可（即使线段通过内部线段的交点此时不影响结果，同样计算了两次）。步骤，（1）对区域四个边界上的点进行排序（源码的方法比较呆板、代码比较冗余，应该有更好的方法）；（2）对四个边界上，相邻焦点的中点与treasure点做线段并与其他线段判断相交，获得最小交点树的情况，即为此题最优。

3、注意事项：注意计算几何模板的应用，判断线段相交的点是有序的。

4、实现方法:

#include < iostream >
#include < algorithm >
using namespace std;
const double eps = 1e - 10 ;

struct point
{
double x, y;
};

struct Tline
{
point p1,p2;
};

point side[ 4 ][ 35 ],End;
Tline line[ 31 ];
int num[ 4 ],ans,n;

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;
}

bool cmp(point p1,point p2)
{
return (p1.y == p2.y && p1.x < p2.x) || (p1.x == p2.x && p1.y < p2.y);
}

void Solve()
{
int i,j;
memset(num, 0 , sizeof (num));
for (i = 0 ;i < n;i ++ )
{
if (line[i].p1.x == 0 ){
side[ 0 ][num[ 0 ]].x = 0 ;
side[ 0 ][num[ 0 ]].y = line[i].p1.y;
num[ 0 ] ++ ;}
if (line[i].p2.x == 0 ){
side[ 0 ][num[ 0 ]].x = 0 ;
side[ 0 ][num[ 0 ]].y = line[i].p2.y;
num[ 0 ] ++ ;}

if (line[i].p1.x == 100 ){
side[ 1 ][num[ 1 ]].x = 100 ;
side[ 1 ][num[ 1 ]].y = line[i].p1.y;
num[ 1 ] ++ ;}
if (line[i].p2.x == 100 ){
side[ 1 ][num[ 1 ]].x = 100 ;
side[ 1 ][num[ 1 ]].y = line[i].p2.y;
num[ 1 ] ++ ;}

if (line[i].p1.y == 0 ){
side[ 2 ][num[ 2 ]].x = line[i].p1.x;
side[ 2 ][num[ 2 ]].y = 0 ;
num[ 2 ] ++ ;}
if (line[i].p2.y == 0 ){
side[ 2 ][num[ 2 ]].x = line[i].p2.x;
side[ 2 ][num[ 2 ]].y = 0 ;
num[ 2 ] ++ ;}

if (line[i].p1.y == 100 ){
side[ 3 ][num[ 3 ]].x = line[i].p1.x;
side[ 3 ][num[ 3 ]].y = 100 ;
num[ 3 ] ++ ;}
if (line[i].p2.y == 100 ){
side[ 3 ][num[ 3 ]].x = line[i].p2.x;
side[ 3 ][num[ 3 ]].y = 100 ;
num[ 3 ] ++ ;}
}
side[ 0 ][num[ 0 ]].x = 0 ;
side[ 0 ][num[ 0 ]].y = 100 ;
side[ 1 ][num[ 1 ]].x = 100 ;
side[ 1 ][num[ 1 ]].y = 100 ;
side[ 2 ][num[ 2 ]].x = 100 ;
side[ 2 ][num[ 2 ]].y = 0 ;
side[ 3 ][num[ 3 ]].x = 100 ;
side[ 3 ][num[ 3 ]].y = 100 ;

for (i = 0 ;i < 4 ;i ++ )
sort(side[i],side[i] + num[i],cmp);
point p,q,mid;
int m,k;
for (i = 0 ;i < 4 ;i ++ )
{
p.x = 0 ;
p.y = 0 ;
for (j = 0 ;j <= num[i];j ++ )
{
m = 1 ;
q = side[i][j];
mid.x = (q.x + p.x) * 0.5 ;
mid.y = (q.y + p.y) * 0.5 ;
for (k = 0 ;k < n;k ++ )
if (inter(mid,End,line[k].p1,line[k].p2))
m ++ ;
if (m < ans)
ans = m;
p = q;
}
}
}

int main()
{
int i;
memset(side, 0 , sizeof (side));
scanf( " %d " , & n);
for (i = 0 ;i < n;i ++ )
{
scanf( " %lf%lf%lf%lf " , & line[i].p1.x, & line[i].p1.y, & line[i].p2.x, & line[i].p2.y);
}
scanf( " %lf%lf " , & End.x, & End.y);
ans = 99999 ;
Solve();
cout << " Number of doors = " << ans << endl;
return 0 ;
}

 

转载于:https://www.cnblogs.com/yongze103/archive/2010/09/24/1833943.html