Java 8-1282 1277 1268 1259 1257字节
通过所有测试。但是，对于其中的一些，它给出的结果会略有不同（当存在一种以上的最佳方法时，这不是问题）。
对于第四项测试，它给出了以下内容：
RDDDDDLD
代替这个：
RDDDDDDL
对于第五项测试，它给出了以下内容：
LLLLUUULLDDLLLLDLLLLLRRRRRRURRRUURRRRRRRRRRRRRRRDDLLRRUULLUUUUUUURRRRRUURRRDRRRLLLLULLLDDLLLLLLUULLLUDLLLLLULLLRRRRRDRRRRRRDDLLLLLLLLLLLLDDDLLLLLLLDURRRRRRRRDDDDRRRRRRUUUUU
代替这个：
UUULLLLLLDDLLLDLLLLLLRRRRRRRRRUUURRRRRRRRRRRRRRRDDLLRRUULLUUUUUUURRRRRUURRRDRRRLLLLULLLLLDDLLLLUULLLUDLLLLLULLLRRRRRDRRRRRRDDLLLLLLLLLLLLDDDLLLLLLLDURRRRRRRRDDDDRRRRRRUUUUU
高尔夫球版：
import java.util.*;class G{int y,w,h,p;String C="",S,o,v;Map m=new HashMap();String q(int a){return a<1?"":"#"+q(a-1);}public static void main(String[]a)throws Exception{new G(new String(java.nio.file.Files.readAllBytes(new java.io.File(a[0]).toPath())));}G(String a){w=(a+"\n").indexOf(10)+3;String t=q(w)+a.replace("\n","##")+q(w);for(char j=65,k=97;j<91;j++,k++){if(t.indexOf(j)*t.indexOf(k)<0)t=t.replace(j,'#').replace(k,' ');}h=t.length()/--w;S=v=q(w*h);t=g(t);if(t!=v)System.out.print(t);}String g(String t){o=(String)m.get(t);if(o!=null)return o;if(t.indexOf(36)<0){if(S.length()>C.length())S=C;return"";}String d="";int f=t.indexOf(64),M[]=new int[w*h],N[]=new int[w*h];Queue<Integer>s=new ArrayDeque();s.add(f);while(!s.isEmpty()){y=s.poll();int[]P={y+1,y-1,y+w,y-w};for(int v:P){char j=t.replaceAll("[A-Z]","#").charAt(v);if(v!=f&j!=35&(N[v]<1|M[y]+1<M[v])){M[v]=M[y]+1;N[v]=y;s.add(v);if(j>32)d+=j;}}}o=d.chars().distinct().mapToObj(e->{String z="",c=C;for(y=t.indexOf(e);y!=f;y=N[y]){p=y-N[y];z=(p==w?"D":p==-w?"U":p==1?"R":"L")+z;}if(S.length()<=(C+z).length())return v;C+=z;String u=g(t.replace('@',' ').replace((char)e,'@').replace((char)(e-32),' '));C=c;return u==v?v:z+u;}).reduce(v,(a,b)->a.length()<b.length()?a:b);m.put(t,o);return o;}}
非高尔夫版本
特征：
import java.util.*;
/**
* @author Victor Stafusa
*/
class TreasureHunt {
// Note: on normal (non-golfing programming) those variables should have been scoped properly.
// They are instance variables just for golfing purposes.
// On the golfed version, nextCellIndex and waypointCellIndex are the same variable. The same also happens to cachedValue and result. This happens is for golfing purposes.
int nextCellIndex,
width,
height,
waypointCellIndex,
cellIndexDifference;
String previousPath = "",
bestSolutionSoFar,
cachedValue,
result,
failureFlag;
// This should be Map<String, String>, but the generics were omitted for golfing.
// It is needed to avoid recomputing long partial dungeons (i.e. dynamic programming).
Map cachedResults = new HashMap();
// Returns a lot of hashes. Like aLotOfHashes(7) will return "#######".
String aLotOfHashes(int howMany) {
return howMany < 1 ? "" : "#" + aLotOfHashes(howMany - 1);
}
// Here is where our program starts.
public static void main(String[] args) throws Exception {
// Read all the content of the file from args[0] and put it into a string.
// Pass that string as a parameter to the constructor.
// The instance itself is useless - it is just a golfing trick.
new TreasureHunt(new String(java.nio.file.Files.readAllBytes(new java.io.File(args[0]).toPath())));
}
// Pre-processs the source in order to format it in the way that we want:
// * No separators between rows. It uses the (row * width + column) formula, so no separators are needed.
// * An extra layer of wall is added in all sides. This naturally fix up problems about walking out of the edges of the board, wrapping-around or acessing invalid array indexes.
// This is a constructor just for golfing purposes. Its instances are worthless.
TreasureHunt(String originalSource) {
// Finds the width by searching the first line-feed.
// If there is just one line and no line-feed, the [+ "\n"] will ensure that it will not break.
// The [+ 3] is because we will add a layer of walls around, so it will be widen by one cell in the left and one in the right (which is +2).
// We still get one more in the width that will be decremented later to use that in the aLotOfHashes method below.
// 10 == '\n'.
width = (originalSource + "\n").indexOf(10) + 3;
// Add a layer of walls in the top and in the bottom (using a lot of hashes for that).
// Replaces the line-feed by a pair of walls, representing the rightmost wall of a row and the leftmost row of the following row.
// Since there is no line-feed before the first line nor after the last line, we add more two walls to fill those.
String newSource = aLotOfHashes(width) + originalSource.replace("\n", "##") + aLotOfHashes(width);
// Remove the keys without door (replaces them as blank spaces) and the doors without keys (replaces them with walls.
// This way, the resulting dungeon will always have matching keys and doors.
// 65 == 'A', 97 == 'a', 91 == 'z'+1
for (char door = 65, key = 97; door < 91; door++, key++) {
// Now a little math trick. For each key or door, we find an index. If the key or door exist, it will be a positive number. Otherwise it will be negative.
// The result will never be zero, because the zeroey position is filled with part of the layer of wall that we added.
// If only one of the key and the door exist, the multiplication will be the product of two numbers with opposite signals, i.e. a negative number.
// Otherwise (both exists or both don't), then the product will be positive.
// So, if the product is negative, we just remove the key and the door (only one of them will be removed of course, but we don't need to care about which one).
if (newSource.indexOf(door) * newSource.indexOf(key) < 0) {
newSource = newSource.replace(door, '#').replace(key, ' ');
}
}
// Knowing the source length and the width (which we fix now), we can easily find out the height.
height = newSource.length() / --width;
// Creates a special value for signaling a non-existence of a path. Since they are sorted by length, this must be a sufficiently large string to always be unfavoured.
bestSolutionSoFar = failureFlag = aLotOfHashes(width * height);
// Now, do the hard work to solve the dungeon...
// Note: On the golfed version, newSource and solution are the same variable.
String solution = solvingRound(newSource);
// If a solution is found, then show it. Otherwise, we just finish without printing anything.
// Note: It is unsafe and a bad practice to compare strings in java using == or != instead of the equals method. However, this code manages the trickery.
if (solution != failureFlag) System.out.print(solution);
}
// This does the hard work, finding a solution for a specific dungeon. This is recursive, so the solution of a dungeon involves the partial solution of the dungeon partially solved.
String solvingRound(String dungeon) {
// To avoid many redundant computations, check if this particular dungeon was already solved before. If it was, return its cached solution.
cachedValue = (String) cachedResults.get(dungeon);
if (cachedValue != null) return cachedValue;
// If there is no treasure in the dungeon (36 == '$'), this should be because the adventurer reached it, so there is no further moves.
if (dungeon.indexOf(36) < 0) {
if (bestSolutionSoFar.length() > previousPath.length()) bestSolutionSoFar = previousPath;
return "";
}
String keysOrTreasureFound = ""; // Initially, we didn't found anything useful.
int adventurerSpot = dungeon.indexOf(64), // 64 == '@', find the cell index of the adventurer.
cellDistance[] = new int[width * height],
previousWaypoint[] = new int[width * height];
// Use a queue to enqueue cell indexes in order to floodfill all the reachable area starting from the adventurer. Again, screw up the proper user of generics.
Queue<Integer> floodFillQueue = new ArrayDeque();
floodFillQueue.add(adventurerSpot); // Seed the queue with the adventurer himself.
// Each cell thies to populate its neighbours to the queue. However no cell will enter the queue more than once if it is not featuring a better path than before.
// This way, all the reachable cells will be reached eventually.
while (!floodFillQueue.isEmpty()) {
nextCellIndex = floodFillQueue.poll();
// Locate the four neighbours of this cell.
// We don't need to bother of checking for wrapping-around or walking into an invalid cell indexes because we added a layer of walls in the beggining,
// and this layer of wall will ensure that there is always something in each direction from any reachable cell.
int[] neighbourCells = {nextCellIndex + 1, nextCellIndex - 1, nextCellIndex + width, nextCellIndex - width};
// For each neighbouring cell...
for (int neighbourCellIndex : neighbourCells) {
// Find the cell content. Considers doors as walls.
char neighbourCellContent = dungeon.replaceAll("[A-Z]", "#").charAt(neighbourCellIndex);
if (neighbourCellIndex != adventurerSpot // If we are not going back to the start ...
& neighbourCellContent != 35 // ... nor walking into a wall or a door that can't be opened (35 == '#') ...
& (previousWaypoint[neighbourCellIndex] < 1 // ... and the neighbour cell is either unvisited ...
| cellDistance[nextCellIndex] + 1 < cellDistance[neighbourCellIndex])) // ... or it was visited before but now we found a better path ...
{ // ... then:
cellDistance[neighbourCellIndex] = cellDistance[nextCellIndex] + 1; // Update the cell distance.
previousWaypoint[neighbourCellIndex] = nextCellIndex; // Update the waypoint so we can track the way from this cell back to the adventurer.
floodFillQueue.add(neighbourCellIndex); // Enqueue the cell once again.
if (neighbourCellContent > 32) keysOrTreasureFound += neighbourCellContent; // If we found something in this cell (32 == space), take a note about that.
}
}
}
// Brute force solutions chosing each one of the interesting things that we found and recursively solving the problem as going to that interesting thing.
// Warning: This has an exponential complexity. Also, if we found something interesting by more than one path, it will compute that redundantly.
result = keysOrTreasureFound.chars().distinct().mapToObj(keyOrTreasure -> {
String tracingWay = "", savedPreviousPath = previousPath;
// From our keyOrTreasure, trace back the path until the adventurer is reached, adding (in reverse order) the steps needed to reach it.
for (waypointCellIndex = dungeon.indexOf(keyOrTreasure); waypointCellIndex != adventurerSpot; waypointCellIndex = previousWaypoint[waypointCellIndex]) {
// Use the difference in cell indexes to see if it is going up, down, right or left.
cellIndexDifference = waypointCellIndex - previousWaypoint[waypointCellIndex];
tracingWay = (cellIndexDifference == width ? "D" : cellIndexDifference == -width ? "U" : cellIndexDifference == 1 ? "R" : "L") + tracingWay;
}
// If this path is going to surely be longer than some other path already found before, prune the search and fail this path.
if (bestSolutionSoFar.length() <= (previousPath + tracingWay).length()) return failureFlag;
// Prepare for recursion, recording the current path as part of the next level recursion's previous path.
previousPath += tracingWay;
// Now that we traced our way from the adventurer to something interesting, we need to continue our jorney through the remaining items.
// For that, create a copy of the dungeon, delete the door of the key that we found (if it was a key),
// move the adventurer to the thing that we just found and recursively solve the resulting simpler problem.
String nextRoundPartialSolution = solvingRound(dungeon
.replace('@', ' ') // Remove the adventurer from where he was...
.replace((char) keyOrTreasure, '@') // ... and put him in the spot of the key or treasure.
.replace((char) (keyOrTreasure - 32), ' ')); // ... and if it was a key, delete the corresponding door ([- 32] converts lowercase to uppercase, won't do anything in the case of the treasure).
// Recursion finished. Now, get back the previous path of the previous recursion level.
previousPath = savedPreviousPath;
// If the subproblem resulted in a failure, then it is unsolvable. Otherwise, concatenates the subproblem solution to the steps that we took.
return nextRoundPartialSolution == failureFlag ? failureFlag : tracingWay + nextRoundPartialSolution;
// From all the paths we took, choose the shorter one.
}).reduce(failureFlag, (a, b) -> a.length() < b.length() ? a : b);
// Now that we have the result of this recursion level and solved this particular dungeon instance,
// cache it to avoid recomputing it all again if the same instance of the dungeon is produced again.
cachedResults.put(dungeon, result);
return result;
}
}
接受输入
要运行它，请尝试以下操作：
javac G.java
java G ./path/to/file/with/dungeon.txt
或者，如果你正在运行的ungolfed版本，替换G用的TreasureHunt。
该文件应包含地牢。输入应不与换行结束。此外，它仅接受\n格式的行尾。不能与\r\n配合使用\r。
而且，它不会验证或清除输入。如果输入格式错误，则行为是不确定的（可能引发异常）。如果找不到该文件，则将引发异常。
备注
我在1100个字节附近的第一个实现无法在合理的时间内解决第5个测试用例。这样做的原因是因为我的实现将所有可获取物品（即未锁在无法进入房间中）的可收集物品（即钥匙和宝藏）的所有可能排列进行了暴力破解。
在最坏的情况下，拥有所有26个键和宝藏，那就是27个！= 10,888,869,450,418,352,160,768,000,000个不同的排列。
OP没有指定答案应该在合理的时间内运行。但是，我认为这是我不希望利用的漏洞。因此，我决定让它在所有测试用例中都可以在可接受的时间内运行。为了实现这一目标，我修改后的程序在搜索路径中进行了修剪，事实证明它比某些已知解决方案更糟糕。此外，它还缓存子解决方案（即动态编程），以避免重新计算可能出现的许多相同的地牢。这样一来，它就可以在我的计算机中仅用一分钟的时间解决第五个测试用例。
解决方案是递归的。这个想法首先是让冒险者去购买某种物品（钥匙或宝藏）。如果有钥匙，冒险者到达它之后，会生成一个新的类似地牢，同时删除钥匙和门，冒险者将移动到钥匙所在的位置。这样，就可以递归求解生成的简单地牢，直到到达宝藏或算法得出结论认为没有任何可到达的物品为止。如上所述，要访问的项目顺序是通过修剪和缓存强制执行的。
冒险者和物品之间的寻路是通过类似于Floodfill和Dijkstra的算法进行的。
最后，我怀疑这个问题是NP完全的（嗯，它的广义版本没有门/钥匙数量的限制）。如果是这样，不要指望能在合理的时间内最佳地解决带有门和钥匙的miryad的大型地牢的解决方案。如果允许使用非最佳路径，那么通过一些启发式方法就很容易处理（如果可能的话就去珍宝，否则去最近的钥匙）。