这几天,准确来说是连续4天了
真的能称之为极限编程了
关于N皇后算法的极限挑战,最终很满意
代码使用了“一维棋盘”,“对称剪枝”,“递归回溯”,“多线程”等特色
最终结果:
15皇后,用时:4903毫秒,计算结果:2279184
16皇后,用时:33265毫秒,计算结果:14772512
17皇后,用时:267460毫秒,计算结果:95815104
比起我第一天写N皇后,14皇后用时87秒的成绩,提高太多了!!!
说好的一定要在100秒内解16皇后,终于解脱了
啥都不说了,贴上代码和运算成绩
package com.newflypig.eightqueen;
import java.util.ArrayList;
import java.util.Date;
import java.util.List;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
public class EightQueen7 {
private static final short K=8; //使用常量来定义,方便之后解N皇后问题
private static short N=0;
public static void main(String[] args) throws Exception {
for(N=9;N<=17;N++){
long count=0;
Date begin =new Date();
/**
* 初始化棋盘,使用一维数组存放棋盘信息
* chess[n]=X:表示第n行X列有一个皇后
*/
List<short[]> chessList=new ArrayList<short[]>(N);
for(short i=0;i<N;i++){
short chess[]=new short[N];
chess[0]=i;
chessList.add(chess);
}
short taskSize =(short)( N/2+(N%2==1?1:0) );
// 创建一个线程池
ExecutorService pool = Executors.newFixedThreadPool(taskSize);
// 创建多个有返回值的任务
List<Future<Long>> futureList = new ArrayList<Future<Long>>(taskSize);
for (int i = 0; i < taskSize; i++) {
Callable<Long> c = new EightQueenThread(chessList.get(i));
// 执行任务并获取Future对象
Future<Long> f = pool.submit(c);
futureList.add(f);
}
// 关闭线程池
pool.shutdown();
for(short i=0; i<(short) (taskSize - (N%2==1?1:0)); i++){
count+=futureList.get(i).get();
}
count=count*2;
if(N%2==1)
count+=futureList.get(N/2).get();
Date end =new Date();
System.out.println("解决 " +N+ "皇后问题,用时:" +String.valueOf(end.getTime()-begin.getTime())+ "毫秒,计算结果:"+count);
}
}
}
class EightQueenThread implements Callable<Long>{
private short[] chess;
private short N;
public EightQueenThread(short[] chess){
this.chess=chess;
this.N=(short) chess.length;
}
@Override
public Long call() throws Exception {
return putQueenAtRow(chess, (short)1) ;
}
private Long putQueenAtRow(short[] chess, short row) {
if(row==N){
return (long) 1;
}
short[] chessTemp=chess.clone();
long sum=0;
/**
* 向这一行的每一个位置尝试排放皇后
* 然后检测状态,如果安全则继续执行递归函数摆放下一行皇后
*/
for(short i=0;i<N;i++){
//摆放这一行的皇后
chessTemp[row]=i;
if( isSafety( chessTemp,row,i) ){
sum+=putQueenAtRow(chessTemp,(short) (row+1));
}
}
return sum;
}
private static boolean isSafety(short[] chess,short row,short col) {
//判断中上、左上、右上是否安全
short step=1;
for(short i=(short) (row-1);i>=0;i--){
if(chess[i]==col) //中上
return false;
if(chess[i]==col-step) //左上
return false;
if(chess[i]==col+step) //右上
return false;
step++;
}
return true;
}
}

这是四天前的成绩:

确实有了很大的提升!