最短路径问题——分支限界法
In the N×M grid array, there is a start-cell, name A, and a end-cell, name B. The problem is finding the shortest routing scheme (i.e. the shortest path) from A to B. When wiring, it can only follow a straight line or a right angle, not a diagonal line. Black cells represent blocked squares that cannot be passed. Shown as the follow picture, in the 3×3 grid array, the length of shortest path from A to B is 6. A tow dimension array can represent the grid array, in which, ordinary cell is presented by 0, and black cell is represented by 1.
Input:
Input N and M at first line (no more than 100), next input N×M array, at last input A's row coordinate, A's column coordinate, B's row coordinate, B's column coordinate.
Output:
print out length of the shortest path from A to B. If no path from A to B, then print 0.
Input Example:
Here is a set of inputs. For example:
3 3
0 1 0
0 1 0
0 0 0
0 0 0 2Output Example:
Here is a set of outputs. For example:
61 package work7; 2 3 import java.util.Scanner; 4 5 /** 6 *@author Noble4 7 *@version 2020年12月5日 下午8:46:42 8 *@package work7 9 **/ 10 public class test2 { 11 static int MAX_VALUE = 100; 12 int row;int line; 13 int[][] Figure;int[][] Tags;//记录到走过的点的步数 14 //移动偏向(-1向左，1向右) 15 int DIRECTION; 16 //起点和终点的坐标 17 int x1;int y1;int x2;int y2; 18 19 20 21 public void Init() { 22 Scanner sr = new Scanner(System.in); 23 row = sr.nextInt(); 24 line = sr.nextInt(); 25 Figure = new int[row][line]; 26 for (int i = 0; i < row; i++) { 27 for (int j = 0; j < line; j++) { 28 Figure[i][j] = sr.nextInt(); 29 } 30 } 31 x1 = sr.nextInt(); 32 y1 = sr.nextInt(); 33 x2 = sr.nextInt(); 34 y2 = sr.nextInt(); 35 Tags = new int[row][line]; 36 //初始化记录步数的数组 37 for (int i = 0; i < row; i++) { 38 for (int j = 0; j < line; j++) { 39 //MAX_VALUE代表没走过 40 Tags[i][j] = MAX_VALUE; 41 } 42 } 43 //初始化移动偏向 44 if(y1 < y2) { 45 DIRECTION = 1; 46 }else if(y1 > y2){ 47 DIRECTION = -1; 48 }else { 49 DIRECTION = 0; 50 } 51 Figure[x1][y1] = 1; 52 Tags[x1][y1] = 0; 53 } 54 55 //单点向周围分支 56 public void Branch(int x,int y,int D) { 57 //限界 58 if(Tags[x][y]>=D) { 59 Tags[x][y] = D; 60 }else { 61 return; 62 } 63 //使用移动偏好 64 if(y <= y1 && DIRECTION == 1) { 65 //越界判断应该在前面不然后面的Figure[x+1][y] == 0可能报错 66 if(!isBorder(x+1, y) && Figure[x+1][y] == 0) { 67 Branch(x+1,y,D+1); 68 } 69 if( !isBorder(x, y+1) && Figure[x][y+1] == 0) { 70 Branch(x,y+1,D+1); 71 } 72 if(!isBorder(x-1, y) && Figure[x-1][y] == 0) { 73 Branch(x-1,y,D+1); 74 } 75 }else if(y >= y1 && DIRECTION == -1) { 76 if(!isBorder(x-1, y) && Figure[x-1][y] == 0) { 77 Branch(x-1,y,D+1); 78 } 79 if(!isBorder(x+1, y) && Figure[x+1][y] == 0) { 80 Branch(x+1,y,D+1); 81 } 82 if(!isBorder(x, y-1) && Figure[x][y-1] == 0) { 83 Branch(x,y-1,D+1); 84 } 85 }else { 86 if(!isBorder(x+1, y) && Figure[x+1][y] == 0) { 87 Branch(x+1,y,D+1); 88 } 89 if(!isBorder(x-1, y) && Figure[x-1][y] == 0) { 90 Branch(x-1,y,D+1); 91 } 92 if(!isBorder(x, y+1) && Figure[x][y+1] == 0) { 93 Branch(x,y+1,D+1); 94 } 95 if(!isBorder(x, y-1) && Figure[x][y-1] == 0) { 96 Branch(x,y-1,D+1); 97 } 98 } 99 } 100 101 //判断是否超过边界 102 public boolean isBorder(int x,int y) { 103 if(x<0||y<0||x >= row||y >= line) { 104 return true; 105 }else { 106 return false; 107 } 108 } 109 110 public static void main(String[] args) { 111 test2 test2 = new test2(); 112 test2.Init(); 113 test2.Branch(test2.x1, test2.y1, 0); 114 System.out.println(test2.Tags[test2.x2][test2.y2]); 115 for (int[] ints : test2.Tags) { 116 for (int item : ints) { 117 System.out.print(item + " "); 118 } 119 System.out.println(); 120 } 121 } 122 }
用例：
7 7 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 4 6
结果截图：
问题：
- 尝试了移动偏好，使用中存在问题，某些特殊情况会得不到最优路径
- 不能在找到最优路径后停止所有寻找
希望有朋友指点迷津！
