जावा 8 - 1282 1277 1268 1259 1257 बाइट्स
यह सभी परीक्षण पास करता है। हालांकि, उनमें से कुछ के लिए यह कुछ अलग परिणाम देता है (जब एक से अधिक इष्टतम तरीका होता है, जो कोई समस्या नहीं है)।
4 वें परीक्षण के लिए, यह यह देता है:
RDDDDDLD
इसके अलावा:
RDDDDDDL
5 वीं परीक्षा के लिए, यह यह देता है:
LLLLUUULLDDLLLLDLLLLLRRRRRRURRRUURRRRRRRRRRRRRRRDDLLRRUULLUUUUUUURRRRRUURRRDRRRLLLLULLLDDLLLLLLUULLLUDLLLLLULLLRRRRRDRRRRRRDDLLLLLLLLLLLLDDDLLLLLLLDURRRRRRRRDDDDRRRRRRUUUUU
इसके अलावा:
UUULLLLLLDDLLLDLLLLLLRRRRRRRRRUUURRRRRRRRRRRRRRRDDLLRRUULLUUUUUUURRRRRUURRRDRRRLLLLULLLLLDDLLLLUULLLUDLLLLLULLLRRRRRDRRRRRRDDLLLLLLLLLLLLDDDLLLLLLLDURRRRRRRRDDDDRRRRRRUUUUU
गोल्फ संस्करण:
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;}}
अनप्लग्ड संस्करण
विशेषताएं:
- जानकारीपूर्ण चर नाम;
- व्याख्यात्मक और विस्तृत टिप्पणियां;
- उचित पहचान।
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;
}
}
इनपुट लेना
इसे चलाने के लिए, इसे आज़माएँ:
javac G.java
java G ./path/to/file/with/dungeon.txt
या फिर आप ungolfed संस्करण चला रहे हैं, की जगह Gके साथ की TreasureHunt।
फ़ाइल में कालकोठरी होनी चाहिए। इनपुट लाइन-फीड के साथ समाप्त नहीं होना चाहिए । इसके अलावा, यह केवल \nप्रारूप में लाइन-एंड को स्वीकार करता है । यह साथ \r\nया साथ काम नहीं करेगा \r।
साथ ही, यह इनपुट को मान्य या पवित्र नहीं करता है। यदि इनपुट विकृत है, तो व्यवहार अपरिभाषित है (अपवाद फेंकने की संभावना)। यदि फ़ाइल नहीं मिल सकती है, तो एक अपवाद फेंक दिया जाएगा।
टिप्पणियों
1100 बाइट्स के पास कहीं मेरा पहला कार्यान्वयन उचित समय में 5 वें परीक्षण मामले को हल नहीं कर सका। इसका कारण यह है कि मेरे कार्यान्वयन के कारण यह संग्रहणीय वस्तुओं (यानी किज़ और खजाना) के सभी संभावित क्रमों को भंग कर देता है, जो कि स्वीकार्य हैं (यानी एक दुर्गम कमरे में बंद नहीं)।
सबसे खराब स्थिति में, सभी 26 कुंजी और खजाने के साथ, यह 27 होगा! = 10,888,869,450,418,352,160,768,000,000 अलग-अलग परमिट।
ओपी ने यह निर्दिष्ट नहीं किया कि उत्तर कुछ ऐसा होना चाहिए जो उचित समय में चले। हालांकि, मुझे लगता है कि यह एक खामी है जिसका मैं शोषण नहीं करना चाहूंगा। इसलिए, मैंने इसे सभी परीक्षण मामलों के लिए स्वीकार्य समय में चलाने का फैसला किया। इसे प्राप्त करने के लिए, मेरे संशोधित कार्यक्रम में खोज रास्तों में छंटनी की सुविधा है जो पहले से ही ज्ञात समाधान से भी बदतर साबित होती है। इसके अलावा, यह कई समान काल कोठरी के पुनर्संयोजन से बचने के लिए सदस्यता (अर्थात गतिशील प्रोग्रामिंग) को भी कैश करता है। इसके साथ, यह मेरे कंप्यूटर में केवल एक मिनट में 5 वें परीक्षण मामले को हल करने में सक्षम है।
समाधान पुनरावर्ती है। इस विचार को सबसे पहले एडवेंचरर को किसी वस्तु (एक कुंजी या खज़ाना) में लाना है। एक कुंजी के मामले में, एडवेंचरर के पहुंचने के बाद, कुंजी और डोर दोनों को हटाकर एक समान समान कालकोठरी उत्पन्न की जाती है और एडवेंचर उस स्थान पर चला जाता है जहां कुंजी थी। उस के साथ, उत्पन्न सरल डंगऑन को पुनरावर्ती रूप से हल किया जाता है जब तक कि जब बिंदु तक पहुंच नहीं जाता है या एल्गोरिथ्म निष्कर्ष निकालता है कि कोई भी पहुंच योग्य वस्तु नहीं है। जिन वस्तुओं का दौरा किया जाना है, वे ऊपर बताए अनुसार छंटाई और कैशिंग के साथ जानवर-मजबूर हैं।
एडवेंचरर और आइटम के बीच का पथ-पता एक एल्गोरिथ्म के साथ बनाया गया है जो फ्लडफिल और डीजकस्ट्रा दोनों से मिलता जुलता है।
अंत में, मुझे संदेह है कि यह समस्या एनपी-पूर्ण है (ठीक है, दरवाजे या चाबियों की संख्या पर सीमा के बिना इसका सामान्यीकृत संस्करण)। यदि यह सच है, तो ऐसे समाधानों की अपेक्षा न करें जो उचित समय में दरवाजों और चाबियों के मिरियड के साथ बहुत बड़े काल कोठरी को हल करते हैं। यदि उप-इष्टतम रास्तों की अनुमति दी गई थी, तो यह आसानी से कुछ अनुमानों के साथ ट्रैक्टेबल होगा (यदि संभव हो तो खजाने पर जाएं, अन्यथा निकटतम कुंजी पर जाएं)।