我正在尝试解决在NxN板上放置N个皇后而没有行，列和对角线冲突的问题。我使用最小化冲突的算法。首先，在每列上随机放置一个女王。之后，从所有冲突女王中随机选择一个，并为她的列计算每个可能位置的冲突。然后，女王以最少的冲突次数移动到最佳位置。它可以工作，但运行速度非常慢。我的目标是使其快速运行10000个皇后。您能否请我建议一些改进或在逻辑上发现一些错误？
这是我的代码：
public class Queen {
int column;
int row;
int d1;
int d2;
public Queen(int column, int row, int d1, int d2) {
super();
this.column = column;
this.row = row;
this.d1 = d1;
this.d2 = d2;
}
@Override
public String toString() {
return "Queen [column=" + column + ", row=" + row + ", d1=" + d1
+ ", d2=" + d2 + "]";
}
@Override
public boolean equals(Object obj) {
return ((Queen)obj).column == this.column && ((Queen)obj).row == this.row;
}
}
和：
import java.util.HashSet;
import java.util.Random;
public class SolveQueens {
public static boolean printBoard = false;
public static int N = 100;
public static int maxSteps = 2000000;
public static int[] queens = new int[N];
public static Random random = new Random();
public static HashSet<Queen> q = new HashSet<Queen>();
public static HashSet rowConfl[] = new HashSet[N];
public static HashSet d1Confl[] = new HashSet[2*N - 1];
public static HashSet d2Confl[] = new HashSet[2*N - 1];
public static void init () {
int r;
rowConfl = new HashSet[N];
d1Confl = new HashSet[2*N - 1];
d2Confl = new HashSet[2*N - 1];
for (int i = 0; i < N; i++) {
r = random.nextInt(N);
queens[i] = r;
Queen k = new Queen(i, r, i + r, N - 1 + i - r);
q.add(k);
if (rowConfl[k.row] == null) {
rowConfl[k.row] = new HashSet<Queen>();
}
if (d1Confl[k.d1] == null) {
d1Confl[k.d1] = new HashSet<Queen>();
}
if (d2Confl[k.d2] == null) {
d2Confl[k.d2] = new HashSet<Queen>();
}
((HashSet<Queen>)rowConfl[k.row]).add(k);
((HashSet<Queen>)d1Confl[k.d1]).add(k);
((HashSet<Queen>)d2Confl[k.d2]).add(k);
}
}
public static void print () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(queens[i] == j ? "♕ " : "◻◻◻ ");
}
System.out.println();
}
System.out.println();
}
public static boolean checkItLinear() {
Queen r = choseConflictQueen();
if (r == null) {
return true;
}
Queen newQ = findNewBestPosition(r);
q.remove(r);
q.add(newQ);
rowConfl[r.row].remove(r);
d1Confl[r.d1].remove(r);
d2Confl[r.d2].remove(r);
if (rowConfl[newQ.row] == null) {
rowConfl[newQ.row] = new HashSet<Queen>();
}
if (d1Confl[newQ.d1] == null) {
d1Confl[newQ.d1] = new HashSet<Queen>();
}
if (d2Confl[newQ.d2] == null) {
d2Confl[newQ.d2] = new HashSet<Queen>();
}
((HashSet<Queen>)rowConfl[newQ.row]).add(newQ);
((HashSet<Queen>)d1Confl[newQ.d1]).add(newQ);
((HashSet<Queen>)d2Confl[newQ.d2]).add(newQ);
queens[r.column] = newQ.row;
return false;
}
public static Queen choseConflictQueen () {
HashSet<Queen> conflictSet = new HashSet<Queen>();
boolean hasConflicts = false;
for (int i = 0; i < 2*N - 1; i++) {
if (i < N && rowConfl[i] != null) {
hasConflicts = hasConflicts || rowConfl[i].size() > 1;
conflictSet.addAll(rowConfl[i]);
}
if (d1Confl[i] != null) {
hasConflicts = hasConflicts || d1Confl[i].size() > 1;
conflictSet.addAll(d1Confl[i]);
}
if (d2Confl[i] != null) {
hasConflicts = hasConflicts || d2Confl[i].size() > 1;
conflictSet.addAll(d2Confl[i]);
}
}
if (hasConflicts) {
int c = random.nextInt(conflictSet.size());
return (Queen) conflictSet.toArray()[c];
}
return null;
}
public static Queen findNewBestPosition(Queen old) {
int[] row = new int[N];
int min = Integer.MAX_VALUE;
int minInd = old.row;
for (int i = 0; i < N; i++) {
if (rowConfl[i] != null) {
row[i] = rowConfl[i].size();
}
if (d1Confl[old.column + i] != null) {
row[i] += d1Confl[old.column + i].size();
}
if (d2Confl[N - 1 + old.column - i] != null) {
row[i] += d2Confl[N - 1 + old.column - i].size();
}
if (i == old.row) {
row[i] = row[i] - 3;
}
if (row[i] <= min && i != minInd) {
min = row[i];
minInd = i;
}
}
return new Queen(old.column, minInd, old.column + minInd, N - 1 + old.column - minInd);
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
init();
int steps = 0;
while(!checkItLinear()) {
if (++steps > maxSteps) {
init();
steps = 0;
}
}
long endTime = System.currentTimeMillis();
System.out.println("Done for " + (endTime - startTime) + "ms\n");
if(printBoard){
print();
}
}
}
编辑：
这是我的一点点优化的解决方案，它删除了一些未使用的对象，并在初始化时将皇后区放在对角线位置。
import java.util.Random;
import java.util.Vector;
public class SolveQueens {
public static boolean PRINT_BOARD = true;
public static int N = 10;
public static int MAX_STEPS = 5000;
public static int[] queens = new int[N];
public static Random random = new Random();
public static int[] rowConfl = new int[N];
public static int[] d1Confl = new int[2*N - 1];
public static int[] d2Confl = new int[2*N - 1];
public static Vector<Integer> conflicts = new Vector<Integer>();
public static void init () {
random = new Random();
for (int i = 0; i < N; i++) {
queens[i] = i;
}
}
public static int getD1Pos (int col, int row) {
return col + row;
}
public static int getD2Pos (int col, int row) {
return N - 1 + col - row;
}
public static void print () {
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
System.out.print(queens[i] == j ? "Q " : "* ");
}
System.out.println();
}
System.out.println();
}
public static boolean hasConflicts() {
generateConflicts();
if (conflicts.isEmpty()) {
return false;
}
int r = random.nextInt(conflicts.size());
int conflQueenCol = conflicts.get(r);
int currentRow = queens[conflQueenCol];
int bestRow = currentRow;
int minConfl = getConflicts(conflQueenCol, queens[conflQueenCol]) - 3;
int tempConflCount;
for (int i = 0; i < N ; i++) {
tempConflCount = getConflicts(conflQueenCol, i);
if (i != currentRow && tempConflCount <= minConfl) {
minConfl = tempConflCount;
bestRow = i;
}
}
queens[conflQueenCol] = bestRow;
return true;
}
public static void generateConflicts () {
conflicts = new Vector<Integer>();
rowConfl = new int[N];
d1Confl = new int[2*N - 1];
d2Confl = new int[2*N - 1];
for (int i = 0; i < N; i++) {
int r = queens[i];
rowConfl[r]++;
d1Confl[getD1Pos(i, r)]++;
d2Confl[getD2Pos(i, r)]++;
}
for (int i = 0; i < N; i++) {
int conflictsCount = getConflicts(i, queens[i]) - 3;
if (conflictsCount > 0) {
conflicts.add(i);
}
}
}
public static int getConflicts(int col, int row) {
return rowConfl[row] + d1Confl[getD1Pos(col, row)] + d2Confl[getD2Pos(col, row)];
}
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
init();
int steps = 0;
while(hasConflicts()) {
if (++steps > MAX_STEPS) {
init();
steps = 0;
}
}
long endTime = System.currentTimeMillis();
System.out.println("Done for " + (endTime - startTime) + "ms\n");
if(PRINT_BOARD){
print();
}
}
}
最佳答案
评论会很有帮助:)
除了重新创建冲突集和“最糟糕的冲突”之后，您还可以创建它一次，然后仅更新已更改的行/列吗？
编辑0：
我尝试了一下您的代码。由于代码是随机的，因此很难确定更改是否正确，因为您可能会以良好的初始状态或糟糕的状态开始。我尝试用10个皇后进行10次跑步，并得出了截然不同的答案，但结果如下。
我伪装分析了哪些语句执行得最多，结果表明selectConflictQueen中的内部循环语句执行得最多。我尝试插入一个中断以拉出第一个冲突女王（如果找到），但似乎并没有太大帮助。
仅对耗时超过一秒的运行进行分组：
我意识到我只有10次跑步，这在统计上还算是不够的，但是，嘿。
因此，增加休息时间似乎无济于事。我认为建设性的解决方案可能会更快，但是随机性将再次使检查变得更加困难。
关于java - 优化N皇后拼图，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/8022278/