具有障碍物的欧几里德最短路径问题及其实现撰文：曾毅
最后更新：2004年3月13日
声明：文章由我执笔整理，程序代码由Kenji Ikeda写于1997年
两个有障碍物地点之间的最短路径问题。不是一个新问题，很久以前就有人提出并解决了这个问题。在历史上这个问题是计算几何学中的一个经典课题。被称为具有障碍物的欧几里德最短路径问题(ESP0)。ESPO可以描述如下:给定平面中两点s和t，及多边形障碍物集Ω={ω1,ω2,...,ωk}，要求由s至t并避开所有障碍物的最短路径。
平面中的ESPO问题在多项式时间内可以求解，而E^3中具有多面体障碍物时确定最短路径长度的问题是NP难的。
一个简单的算法如下：由s到t的最短路径是一条折线链，该链的顶点是多边形障碍物的顶点。利用Dijkstra的最短路算法和可视图算法可以求解ESPO问题，其时间复杂性为0(n^2)。如果可视图是稀疏的，则在O(m+nlogn)时间内可以确定解，其中m是可视图边的数目。
下面结合最短路径问题简单介绍一下Dijkstra算法以便大家学习：
最短路径问题：
在联结图G=(V,E)中，顶点集E->R+(即是权值为正) ，在点集V中的固定顶点s，找s到V中各顶点 v的最短路径.
Dijkstra算法Dijkstra算法是一种解决最短路径问题的非常有效的算法，时间复杂度为 O(|V|2):(这段是MIT一位老先生写得，不翻译了，保持原作)
QUOTE
1. Set i=0, S0= {u0=s}, L(u0)=0, and L(v)=infinity for v <> u0. If |V| = 1 then stop, otherwise go to step 2.
2. For each v in V\Si, replace L(v) by min{L(v), L(ui)+dvui}. If L(v) is replaced, put a label (L(v), ui) on v. 3. Find a vertex v which minimizes {L(v): v in V\Si}, say ui+1.
4. Let Si+1 = Si cup {ui+1}.
5. Replace i by i+1. If i=|V|-1 then stop, otherwise go to step 2.
6. Set i=0, S0= {u0=s}, L(u0)=0, and L(v)=infinity for v <> u0. If |V| = 1 then stop, otherwise go to step 2.
7. For each v in V\Si, replace L(v) by min{L(v), L(ui)+dvui}. If L(v) is replaced, put a label (L(v), ui) on v. 8. Find a vertex v which minimizes {L(v): v in V\Si}, say ui+1.
9. Let Si+1 = Si cup {ui+1}.
10. Replace i by i+1. If i=|V|-1 then stop, otherwise go to step 2.
下面给出Kenji Ikeda老前辈的实现程序(Java版)：
CODE
/********************************************/
/* Dijkstra.java */
/* Copyright (C) 1997, 1998, 2000 K. Ikeda */
/********************************************/
import java.applet.*;
import java.awt.*;
import java.awt.event.*;
import java.util.*;
import java.io.*;
import java.net.URL;
class Node {
int x;
int y;
int delta_plus; /* edge starts from this node */
int delta_minus; /* edge terminates at this node */
int dist; /* distance from the start node */
int prev; /* previous node of the shortest path */
int succ,pred; /* node in Sbar with finite dist. */
int w;
int h;
int pw;
int dx;
int dy;
String name;
}
class Edge {
int rndd_plus; /* initial vertex of this edge */
int rndd_minus; /* terminal vertex of this edge */
int delta_plus; /* edge starts from rndd_plus */
int delta_minus; /* edge terminates at rndd_minus */
int len; /* length */
String name;
}
public class Dijkstra extends Applet implements MouseListener {
int n,m;
int u,snode; /* start node */
int pre_s_first, pre_s_last;
boolean isdigraph;
int iteration, step;
Node v[] = new Node[100];
Edge e[] = new Edge[200];
int findNode(String name) {
for (int i=0; i<n; i++)
if (v[i].name.equals(name))
return i;
return -1;
}
void input_graph(InputStream is) throws IOException {
int x,y,l;
String s;
Reader r = new BufferedReader(new InputStreamReader(is));
StreamTokenizer st = new StreamTokenizer(r);
st.commentChar('#');
st.nextToken(); n = (int)st.nval;
st.nextToken(); m = (int)st.nval;
st.nextToken(); s = st.sval;
isdigraph = "digraph".equals(s);
for (int i = 0; i<n; i++) {
Node node = new Node();
st.nextToken(); node.name = st.sval;
st.nextToken(); node.x = (int)st.nval;
st.nextToken(); node.y = (int)st.nval;
v[i] = node;
}
for (int i = 0; i<m; i++) {
Edge edge = new Edge();
st.nextToken(); edge.name = st.sval;
switch (st.nextToken()) {
case StreamTokenizer.TT_NUMBER:
edge.rndd_plus = (int)st.nval;
break;
case StreamTokenizer.TT_WORD:
edge.rndd_plus = findNode(st.sval);
break;
default:
break;
}
switch (st.nextToken()) {
case StreamTokenizer.TT_NUMBER:
edge.rndd_minus = (int)st.nval;
break;
case StreamTokenizer.TT_WORD:
edge.rndd_minus = findNode(st.sval);
break;
default:
break;
}
st.nextToken(); edge.len = (int)st.nval;
e[i] = edge;
}
for (int i=0; i<n; i++) {
v[i].succ = v[i].pred = -2;
v[i].prev = v[i].dist = -1;
v[i].pw = 0;
}
iteration = step = 0;
}
void rdb() {
int i,k;
for (i=0; i<n; i++)
v[i].delta_plus = v[i].delta_minus = -1;
for (i=0; i<m; i++)
e[i].delta_plus = e[i].delta_minus = -1;
for (i=0; i<m; i++) {
k = e[i].rndd_plus;
if (v[k].delta_plus == -1)
v[k].delta_plus = i;
else {
k = v[k].delta_plus;
while(e[k].delta_plus >= 0)
k = e[k].delta_plus;
e[k].delta_plus = i;
}
k = e[i].rndd_minus;
if (v[k].delta_minus == -1)
v[k].delta_minus = i;
else {
k = v[k].delta_minus;
while(e[k].delta_minus >= 0)
k = e[k].delta_minus;
e[k].delta_minus = i;
}
}
}
void append_pre_s(int i) {
v[i].succ = v[i].pred = -1;
if (pre_s_first<0)
pre_s_first = i;
else
v[pre_s_last].succ = i;
v[i].pred = pre_s_last;
pre_s_last = i;
}
void remove_pre_s(int i) {
int succ = v[i].succ;
int pred = v[i].pred;
if (succ>=0)
v[succ].pred = pred;
else
pre_s_last = pred;
if (pred>=0)
v[pred].succ = succ;
else
pre_s_first = succ;
}
void step1() { /* initialize */
u = snode;
for (int i=0; i<n; i++) {
v[i].succ = v[i].pred = -2;
v[i].prev = v[i].dist = -1;
}
v[u].succ = -3;
v[u].dist = 0;
pre_s_first = pre_s_last = -1;
}
void step2() { /* replace labels */
int i,j;
j = v[u].delta_plus;
while (j>=0) {
i = e[j].rndd_minus;
if ((v[i].succ>=-2)&&((v[i].dist<0)||
(v[i].dist>v[u].dist+e[j].len))) {
if (v[i].dist<0)
append_pre_s(i);
v[i].dist = v[u].dist + e[j].len;
v[i].prev = u; /* label */
}
j = e[j].delta_plus;
}
if (!isdigraph) {
j = v[u].delta_minus;
while (j>=0) {
i = e[j].rndd_plus;
if ((v[i].succ>=-2)&&((v[i].dist<0)||
(v[i].dist>v[u].dist+e[j].len))) {
if (v[i].dist<0)
append_pre_s(i);
v[i].dist = v[u].dist + e[j].len;
v[i].prev = u; /* label */
}
j = e[j].delta_minus;
}
}
v[u].succ = -4;
}
void step3() { /* find the shortest node in Sbar */
int i,rho;
rho = -1;
for (i = pre_s_first; i>=0; i = v[i].succ) {
if ((rho < 0)||(rho>v[i].dist)) {
rho = v[i].dist;
u = i;
}
}
remove_pre_s(u);
v[u].succ = -3;
}
void step4() {
v[u].succ = -4;
}
double weight(Node n, double x, double y) {
double w,z,xx,yy;
w = 0;
for (int j = n.delta_plus; j>=0; j=e[j].delta_plus) {
xx = (double)(v[e[j].rndd_minus].x - n.x);
yy = (double)(v[e[j].rndd_minus].y - n.y);
z = (x*xx+y*yy)/Math.sqrt((x*x+y*y)*(xx*xx+yy*yy))+.0;
w += z*z*z*z;
}
for (int j = n.delta_minus; j>=0; j=e[j].delta_minus) {
xx = (double)(v[e[j].rndd_plus].x - n.x);
yy = (double)(v[e[j].rndd_plus].y - n.y);
z = (x*xx+y*yy)/Math.sqrt((x*x+y*y)*(xx*xx+yy*yy))+1.0;
w += z*z*z*z;
}
return w;
}
void init_sub() {
int x[] = {1, 0, -1, 1, 0, -1};
int y[] = {1, 1, 1, -1, -1, -1};
int i,j,k;
double w,z;
for (i=0; i<n; i++) {
k=0;
w=weight(v[i],(double)x[0],(double)y[0]);
for (j=1; j<6; j++) {
z=weight(v[i],(double)x[j],(double)y[j]);
if (z<w) {
w = z;
k = j;
}
}
v[i].dx = x[k];
v[i].dy = y[k];
}
}
public void init() {
String mdname = getParameter("inputfile");
try {
InputStream is;
is = new URL(getDocumentBase(),mdname).openStream();
input_graph(is);
try {
if (is != null)
is.close();
} catch(Exception e) {
}
} catch (FileNotFoundException e) {
System.err.println("File not found.");
} catch (IOException e) {
System.err.println("Cannot access file.");
}
String s = getParameter("start");
if (s != null)
snode = Integer.parseInt(s);
else
snode = 0;
setBackground(Color.white);
rdb();
init_sub();
addMouseListener(this);
}
public void paintNode(Graphics g, Node n, FontMetrics fm) {
String s;
int x = n.x;
int y = n.y;
int w = fm.stringWidth(n.name) + 10;
int h = fm.getHeight() + 4;
n.w = w;
n.h = h;
if (n.succ<-2)
g.setColor(Color.blue);
else if (n.succ==-2)
g.setColor(Color.gray);
else
g.setColor(Color.red);
g.drawRect(x-w/2,y-h/2,w,h);
if (n.succ==-4)
g.setColor(Color.cyan);
else if (n.succ==-3)
g.setColor(Color.pink);
else if (n.succ>-2)
g.setColor(Color.yellow);
else
g.setColor(getBackground());
g.fillRect(x-w/2+1,y-h/2+1,w-1,h-1);
g.setColor(Color.black);
g.drawString(n.name,x-(w-10)/2,(y-(h-4)/2)+fm.getAscent());
if (n.dist<0)
s = "";
else
s = ""+n.dist;
w = fm.stringWidth(s) + 10;
x += (h+1)*n.dx; y += (h+1)*n.dy;
g.setColor(getBackground());
g.fillRect(x-n.pw/2,y-h/2,n.pw,h);
n.pw = w;
if (n.succ<-2)
g.setColor(Color.blue);
else
g.setColor(Color.red);
g.drawString(s,x-(w-10)/2,y-(h-4)/2+fm.getAscent());
}
int [] xy(int a, int b, int w, int h) {
int x[] = new int[2];
if (Math.abs(w*b)>=Math.abs(h*a)) {
x[0] = ((b>=0)?1:-1)*a*h/b/2;
x[1] = ((b>=0)?1:-1)*h/2;
} else {
x[0] = ((a>=0)?1:-1)*w/2;
x[1] = ((a>=0)?1:-1)*b*w/a/2;
}
return x;
}
void drawArrow(Graphics g,int x1,int y1,int x2,int y2) {
int a = x1-x2;
int b = y1-y2;
if (isdigraph) {
double aa = Math.sqrt(a*a+b*b)/16.0;
double bb = b/aa;
aa = a/aa;
g.drawLine(x2,y2,x2+(int)((aa*12+bb*5)/13),y2+(int)((-aa*5+bb*12)/13));
g.drawLine(x2,y2,x2+(int)((aa*12-bb*5)/13),y2+(int)((aa*5+bb*12)/13));
}
g.drawLine(x1,y1,x2,y2);
}
public void paintEdge(Graphics g, Edge e, FontMetrics fm) {
Node v1 = v[e.rndd_plus];
Node v2 = v[e.rndd_minus];
int a = v1.x-v2.x;
int b = v1.y-v2.y;
int x1[] = xy(-a,-b,v1.w,v1.h);
int x2[] = xy(a,b,v2.w,v2.h);
if (v2.prev == e.rndd_plus) {
if ((v1.succ<-2)&&(v2.succ>=-2))
g.setColor(Color.red);
else
g.setColor(Color.blue);
} else {
g.setColor(Color.lightGray);
}
if ((!isdigraph)&&(v1.prev == e.rndd_minus)) {
if ((v2.succ<-2)&&(v1.succ>=-2))
g.setColor(Color.red);
else
g.setColor(Color.blue);
}
drawArrow(g,v1.x+x1[0],v1.y+x1[1],v2.x+x2[0],v2.y+x2[1]);
int w = fm.stringWidth("" + e.len);
int h = fm.getHeight();
g.setColor(getBackground());
g.fillRect((v1.x+v2.x-w)/2,(v1.y+v2.y-h)/2,w,h);
if ((v2.prev == e.rndd_plus)||
((!isdigraph)&&(v1.prev == e.rndd_minus)))
g.setColor(Color.black);
else
g.setColor(Color.lightGray);
g.drawString("" + e.len,(v1.x+v2.x-w)/2,(v1.y+v2.y-h)/2+fm.getAscent());
}
public void paint(Graphics g) {
FontMetrics fm = g.getFontMetrics();
for (int i=0; i<n; i++)
paintNode(g,v[i],fm);
for (int i=0; i<m; i++)
paintEdge(g,e[i],fm);
}
public void update(Graphics g) {
paint(g);
}
public void mousePressed(MouseEvent ev) {
if (iteration==0) {
step1();
iteration++;
step = 2;
} else if (iteration>=n) {
step4();
iteration = 0;
} else {
if (step == 2) {
step2();
step = 3;
} else {
step3();
iteration++;
step = 2;
}
}
repaint();
}
public void mouseClicked(MouseEvent event) {}
public void mouseReleased(MouseEvent event) {}
public void mouseEntered(MouseEvent event) {}
public void mouseExited(MouseEvent event) {}
}
对于该问题还有一些更好的算法，下面作一些简单的介绍：利用最短路径映射SPM(s,Ω)在O(n(k+logn))时间内求解任意多边形障碍物的ESPO问题的方法是由Reif和Storer提出的。如果给定SPM(s,Ω)，则在O(logn)时间内可以确定包含t的域，而在0(b+logn)时间内能够确定到t的路径，其中b是路径上线段的数目。Welzl等人利用可视图给出了求解平面上n条线段的ESPO问题的算法，该算法要求0(n^2)时间。不难修改这个算法使其能处理多边形障碍物，并且具有相同的时间复杂性。注意，如果使用可视图方法，那么对限界0(n^2)将不可能改进。多边形物体中两个物体(而非点)之间的最短路径的0(n^2)算法是已知的。当n是平行线段集合时，Lee和Preparata提出θ(nlogn)平面扫描算法。线段穿过扫描线并且把最短路径映射到扫描线。平面上没有最短路径的0(n^2)算法能处理避开n条任意相交的线段。Rohnert给出平面中避开k个凸障碍物最短路径的O(nlogn+k^2)时间的算法。这个时间限界在O(k^2logn+n)时间和O(n+k^2)空间预处理障碍物的条件下达到。预处理包括构造可视图的子图。Rohnert还给出平面中避开k个凸障碍物最短路径的O(knlogn)时间和O(n)空间的算法。后者不需预先处理障碍物，而是利用Dijkstra最短路径算法在线计算可视性。当平面中有k个凸障碍物并且其边界至多相交两次时，Rohnert给出的算法能找到平面中任意两点之间的最短路径，其时间复杂性为O(nlogn+k^2)。这个时间限界在O(nlogn+k^3)时间和O(n+k^2)空间预处理障碍物的条件下达到。
