最近我试图解决着名的little bishops算法问题。在我阅读的其中一个网站中,我应该将棋盘分成黑白部分以优化执行。之后,我应该使用回溯来统计可能的方法,以便将主教分成黑色方块和白色方块。回溯优化
在下面的代码中,我尝试将6个主教放在8乘8棋盘的白色方块上。我这样做只是为了验证技术是否真的有效。
//inside main function
int k = 6; //number of bishops
int n = 8; //length of one side of chessboard
Integer[] positions = new Integer[k];
long result = backtrack(positions, 0, n);
//find how many times we double counting each possible combination of bishops
int factor = 1;
for(int i = k; i>0; i--) {
factor = factor * i;
}
System.out.println("The result is " + result/factor);
//implementation of other functions
public long backtrack(Integer[] prevPositions, int k, int n) {
if(k == 6) {
return 1;
}
long sum = 0;
Integer[] candidates = new Integer[n*n];
int length = getCandidates(prevPositions, k, candidates, n);
for(int i=0 ; i<length ; i++) {
prevPositions[k] = candidates[i];
sum += backtrack(prevPositions,k+1,n);
}
return sum;
}
public Integer getCandidates(Integer[] prevPositions, int k, Integer[] candidates, int n) {
int length = 0;
//only white squares are considered as candidates, hence i+=2
for (int i = 0; i < n*n; i+=2) {
boolean isGood = true;
int iRow = i/n;
int iCol = i % n;
for (int j = 0; j < k; j++) {
int prev = prevPositions[j];
if (i == prev) {
isGood = false;
break;
} else {
int prevRow = prev/n;
int prevCol = prev % n;
if (Math.abs(iRow - prevRow) == Math.abs(iCol - prevCol)) {
isGood = false;
break;
}
}
}
if(isGood) {
candidates[length] = new Integer(i);
length++;
}
}
return length;
}
即使我能看到为什么划分成棋盘白色和黑色方块优化问题,它仍然需要大约11秒计数可能的方式把所有的主教只白色方块数。你能帮我吗?我究竟做错了什么?
你看过吗? http://rosettacode.org/wiki/N-queens_problem它能给你提供处理主教的想法吗?也可以看看:http://www.cs.sunysb.edu/~skiena/392/newlectures/week8.pdf它有一个适用于n皇后的回溯应用程序,以及小主教的示例练习。 –