一、深度优先搜索
深度优先搜索就是在搜索树的每一层始终先只扩展一个子节点，不断地向纵深前进直到不能再前进（到达叶子节点或受到深度限制）时，才从当前节点返回到上一级节点，沿另一方向又继续前进。这种方法的搜索树是从树根开始一枝一枝逐渐形成的。
深度优先搜索亦称为纵向搜索。由于一个有解的问题树可能含有无穷分枝，深度优先搜索如果误入无穷分枝（即深度无限），则不可能找到目标节点。所以，深度优先搜索策略是不完备的。另外，应用此策略得到的解不一定是最佳解（最短路径）。
二、 重排九宫问题游戏
在一个3乘3的九宫中有1-8的8个数及一个空格随机摆放在其中的格子里。如下面左图所示。现在要求实现这样的问题：将该九宫调整为如下图右图所示的形式。调整规则是：每次只能将与空格（上，下或左，右）相临的一个数字平移到空格中。试编程实现。
| 2 | 8 | 3 | | 1 | 2 | 3 |
-
| 1 | | 4 | | 8 | | 4 |
| 7 |6 | 5 | | 7 | 6 | 5 |
深度优先搜索的路径示意图：
三、广度优先搜索
在深度优先搜索算法中，是深度越大的结点越先得到扩展。如果在搜索中把算法改为按结点的层次进行搜索， 本层的结点没有搜索处理完时，不能对下层结点进行处理，即深度越小的结点越先得到扩展，也就是说先产生 的结点先得以扩展处理，这种搜索算法称为广度优先搜索法。
广度优先搜索路径示意图：
四、航班问题（来自《The Art of Java》)
一位顾客要预定一张从New York到Los Angeles的航班机票，下面是航班线路，请你为顾客找一种购票方案。
|
航班
|
距离
|
| New York到Chicago |
900英里
|
| Chicago到Denver |
1000英里
|
| New York到Toronto |
500英里
|
| New York到Denver |
1800英里
|
| Toronto到Calgary |
1700英里
|
| Toronto到Los Angeles |
2500英里
|
| Toronto到Chicago |
500英里
|
| Denver到Urbana |
1000英里
|
| Denver到Houston |
1000英里
|
| Houston到Los Angeles |
1500英里
|
| Denver到Los Angeles |
1000英里
|
下面是用深度优先搜索求解的程序:
import java.util. * ;
import java.io. * ;
// Flightinformation.
class FlightInfo{
Stringfrom;
Stringto;
int distance;
boolean skip; // usedinbacktracking
FlightInfo(Stringf,Stringt, int d){
from = f;
to = t;
distance = d;
skip = false ;
}
}
class Depth{
final int MAX = 100 ;
// Thisarrayholdstheflightinformation.
FlightInfoflights[] = new FlightInfo[MAX];
int numFlights = 0 ; // numberofentriesinflightarray
StackbtStack = new Stack(); // backtrackstack
public static void main(Stringargs[])
{
Stringto,from;
Depthob = new Depth();
BufferedReaderbr = new
BufferedReader( new InputStreamReader(System.in));
ob.setup();
try {
System.out.print( " From? " );
from = br.readLine();
System.out.print( " To? " );
to = br.readLine();
ob.isflight(from,to);
if (ob.btStack.size() != 0 )
ob.route(to);
} catch (IOExceptionexc){
System.out.println( " Erroroninput. " );
}
}
// Initializetheflightdatabase.
void setup()
{
addFlight( " NewYork " , " Chicago " , 900 );
addFlight( " Chicago " , " Denver " , 1000 );
addFlight( " NewYork " , " Toronto " , 500 );
addFlight( " NewYork " , " Denver " , 1800 );
addFlight( " Toronto " , " Calgary " , 1700 );
addFlight( " Toronto " , " LosAngeles " , 2500 );
addFlight( " Toronto " , " Chicago " , 500 );
addFlight( " Denver " , " Urbana " , 1000 );
addFlight( " Denver " , " Houston " , 1000 );
addFlight( " Houston " , " LosAngeles " , 1500 );
addFlight( " Denver " , " LosAngeles " , 1000 );
}
// Putflightsintothedatabase.
void addFlight(Stringfrom,Stringto, int dist)
{
if (numFlights < MAX){
flights[numFlights] =
new FlightInfo(from,to,dist);
numFlights ++ ;
}
else System.out.println( " Flightdatabasefull. " );
}
// Showtherouteandtotaldistance.
void route(Stringto)
{
Stackrev = new Stack();
int dist = 0 ;
FlightInfof;
int num = btStack.size();
// Reversethestacktodisplayroute.
for ( int i = 0 ;i < num;i ++ )
rev.push(btStack.pop());
for ( int i = 0 ;i < num;i ++ ){
f = (FlightInfo)rev.pop();
System.out.print(f.from + " to " );
dist += f.distance;
}
System.out.println(to);
System.out.println( " Distanceis " + dist);
}
/* Ifthereisaflightbetweenfromandto,
returnthedistanceofflight;
otherwise,return0. */
int match(Stringfrom,Stringto)
{
for ( int i = numFlights - 1 ;i > - 1 ;i -- ){
if (flights[i].from.equals(from) &&
flights[i].to.equals(to) &&
! flights[i].skip)
{
flights[i].skip = true ; // preventreuse
return flights[i].distance;
}
}
return 0 ; // notfound
}
// Givenfrom,findanyconnection.
FlightInfofind(Stringfrom)
{
for ( int i = 0 ;i < numFlights;i ++ ){
if (flights[i].from.equals(from) &&
! flights[i].skip)
{
FlightInfof = new FlightInfo(flights[i].from,
flights[i].to,
flights[i].distance);
flights[i].skip = true ; // preventreuse
return f;
}
}
return null ;
}
// Determineifthereisaroutebetweenfromandto.
void isflight(Stringfrom,Stringto)
{
int dist;
FlightInfof;
// Seeifatdestination.
dist = match(from,to);
if (dist != 0 ){
btStack.push( new FlightInfo(from,to,dist));
return ;
}
// Tryanotherconnection.
f = find(from);
if (f != null ){
btStack.push( new FlightInfo(from,to,f.distance));
isflight(f.to,to);
}
else if (btStack.size() > 0 ){
// Backtrackandtryanotherconnection.
f = (FlightInfo)btStack.pop();
isflight(f.from,f.to);
}
}
}
解释：isflight()方法用递归方法进行深度优先搜索，它先调用match()方法检查航班的数据库，判断在from和to之间有没有航班可达。如果有，则获取目标信息，并将该线路压入栈中，然后返回（找到一个方案）。否则，就调用find()方法查找from与任意其它城市之间的线路，如果找到一条就返回描述该线路的FlightInfo对象，否则返回null。如果存在这样的一条线路，那么就把该线路保存在f中，并将当前航班信息压到栈的顶部，然后递归调用isflight()方法 ,此时保存在f.to中的城市成为新的出发城市.否则就进行回退,弹出栈顶的第一个节点,然后递归调用isflight()方法。该过程将一直持续到找到目标为止。
程序运行结果：
C:\java>java Depth
From? New York
To? Los Angeles
New York to Chicago to Denver to Los Angeles
Distance is 2900
C:\java>
深度优先搜索能够找到一个解，同时，对于上面这个特定问题，深度优先搜索没有经过回退，一次就找到了一个解；但如果数据的组织方式不同，寻找解时就有可能进行多次回退。因此这个例子的输出并不具有普遍性。而且，在搜索一个很长，但是其中并没有解的分支的时候，深度优先搜索的性能将会很差，在这种情况下，深度优先搜索不仅在搜索这条路径时浪费时间，而且还在向目标的回退中浪费时间。
再看对这个例子使用广度优先搜索的程序：
程序运行结果：
C:\java>java Breadth
From? New York
To? Los Angeles
New York to Toronto to Los Angeles
Distance is 3000
import java.util. * ;
import java.io. * ;
// Flightinformation.
class FlightInfo{
Stringfrom;
Stringto;
int distance;
boolean skip; // usedinbacktracking
FlightInfo(Stringf,Stringt, int d){
from = f;
to = t;
distance = d;
skip = false ;
}
}
class Breadth{
final int MAX = 100 ;
// Thisarrayholdstheflightinformation.
FlightInfoflights[] = new FlightInfo[MAX];
int numFlights = 0 ; // numberofentriesinflightarray
StackbtStack = new Stack(); // backtrackstack
public static void main(Stringargs[])
{
Stringto,from;
Breadthob = new Breadth();
BufferedReaderbr = new
BufferedReader( new InputStreamReader(System.in));
ob.setup();
try {
System.out.print( " From? " );
from = br.readLine();
System.out.print( " To? " );
to = br.readLine();
ob.isflight(from,to);
if (ob.btStack.size() != 0 )
ob.route(to);
} catch (IOExceptionexc){
System.out.println( " Erroroninput. " );
}
}
// Initializetheflightdatabase.
void setup()
{
addFlight( " NewYork " , " Chicago " , 900 );
addFlight( " Chicago " , " Denver " , 1000 );
addFlight( " NewYork " , " Toronto " , 500 );
addFlight( " NewYork " , " Denver " , 1800 );
addFlight( " Toronto " , " Calgary " , 1700 );
addFlight( " Toronto " , " LosAngeles " , 2500 );
addFlight( " Toronto " , " Chicago " , 500 );
addFlight( " Denver " , " Urbana " , 1000 );
addFlight( " Denver " , " Houston " , 1000 );
addFlight( " Houston " , " LosAngeles " , 1500 );
addFlight( " Denver " , " LosAngeles " , 1000 );
}
// Putflightsintothedatabase.
void addFlight(Stringfrom,Stringto, int dist)
{
if (numFlights < MAX){
flights[numFlights] =
new FlightInfo(from,to,dist);
numFlights ++ ;
}
else System.out.println( " Flightdatabasefull. " );
}
// Showtherouteandtotaldistance.
void route(Stringto)
{
Stackrev = new Stack();
int dist = 0 ;
FlightInfof;
int num = btStack.size();
// Reversethestacktodisplayroute.
for ( int i = 0 ;i < num;i ++ )
rev.push(btStack.pop());
for ( int i = 0 ;i < num;i ++ ){
f = (FlightInfo)rev.pop();
System.out.print(f.from + " to " );
dist += f.distance;
}
System.out.println(to);
System.out.println( " Distanceis " + dist);
}
/* Ifthereisaflightbetweenfromandto,
returnthedistanceofflight;
otherwise,return0. */
int match(Stringfrom,Stringto)
{
for ( int i = numFlights - 1 ;i > - 1 ;i -- ){
if (flights[i].from.equals(from) &&
flights[i].to.equals(to) &&
! flights[i].skip)
{
flights[i].skip = true ; // preventreuse
return flights[i].distance;
}
}
return 0 ; // notfound
}
// Givenfrom,findanyconnection.
FlightInfofind(Stringfrom)
{
for ( int i = 0 ;i < numFlights;i ++ ){
if (flights[i].from.equals(from) &&
! flights[i].skip)
{
FlightInfof = new FlightInfo(flights[i].from,
flights[i].to,
flights[i].distance);
flights[i].skip = true ; // preventreuse
return f;
}
}
return null ;
}
/* Determineifthereisaroutebetweenfromandto
usingbreadth-firstsearch. */
void isflight(Stringfrom,Stringto)
{
int dist,dist2;
FlightInfof;
// Thisstackisneededbythebreadth-firstsearch.
StackresetStck = new Stack();
// Seeifatdestination.
dist = match(from,to);
if (dist != 0 ){
btStack.push( new FlightInfo(from,to,dist));
return ;
}
/* Followingisthefirstpartofthebreadth-first
modification.Itchecksallconnectingflights
fromaspecifiednode. */
while ((f = find(from)) != null ){
resetStck.push(f);
if ((dist = match(f.to,to)) != 0 ){
resetStck.push(f.to);
btStack.push( new FlightInfo(from,f.to,f.distance));
btStack.push( new FlightInfo(f.to,to,dist));
return ;
}
}
/* Thefollowingcoderesetstheskipfieldssetby
precedingwhileloop.Thisisalsopartofthe
breadth-firstmodifiction. */
int i = resetStck.size();
for (;i != 0 ;i -- )
resetSkip((FlightInfo)resetStck.pop());
// Tryanotherconnection.
f = find(from);
if (f != null ){
btStack.push( new FlightInfo(from,to,f.distance));
isflight(f.to,to);
}
else if (btStack.size() > 0 ){
// Backtrackandtryanotherconnection.
f = (FlightInfo)btStack.pop();
isflight(f.from,f.to);
}
}
// Resetskipfieldofspecifiedflight.
void resetSkip(FlightInfof){
for ( int i = 0 ;i < numFlights;i ++ )
if (flights[i].from.equals(f.from) &&
flights[i].to.equals(f.to))
flights[i].skip = false ;
}
}
C:\java>
它找到了一个合理的解，但这不具有一般性。因为找到的第一条路径取决于信息的物理组织形式。
如果目标在搜索空间中隐藏得不是太深，那么广度优先搜索的性能会很好。
