最短路径条数问题

问题描述：

给定如图所示的无向连通图，假定图中所有边的权值都为1，显然，从源点A到终点T的虽短路径有多条，求不同的最短路径的数目。

                                      

 

权值相同的最短路径问题，则但愿点Dijkstra算法退化成广度优先搜索，假定起点为0，终点为N。

用动态规划的思想：

• 使用两个辅助数组：
  • 步数：step[0...N]，记录从起点到某个顶点i的走的最小步数；
  • 路径条数：path[0...N]，记录从起点到某个顶点的最短路径的条数；
  • 总体思路是：根据步数更新最短路径的条数。
•  考虑：当从当前顶点i扩散到其某相邻顶点j时：
  • 如果step[j] == 0，说明之前还没有路径到达过顶点j；活着step[j] > step[i] + 1，则说明之前有顶点走过这条边，且步数还要比目前走的路径长，因此可以用当前路径i的信息更新j.
    • 则step[j] = step[i] + 1;
    • path[j] = path[i];
  • 如果step[j] == step[i] + 1，说明之前有顶点走过这条边，并且步数跟从i到j的步数一样，都是最短路径，因此更新j.
    • step[j] 不变;
    • path[j] = path[j] + path[i];
•  用队列保存目前遍历的节点。

 

 Code:

class NumOfShortestPath {
    private int[][] aja = {
            /*0*/{0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0},
            /*1*/{1,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0},
            /*2*/{0,1,0,1,0,0,1,0,0,0,0,0,0,0,0,0},
            /*3*/{0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0},
            /*4*/{1,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0},
            /*5*/{0,1,0,0,1,0,1,0,0,1,0,0,0,0,0,0},
            /*6*/{0,0,1,0,0,1,0,1,0,0,1,0,0,0,0,0},
            /*7*/{0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0},
            /*8*/{0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0},
            /*9*/{0,0,0,0,0,1,0,0,1,0,1,0,0,1,0,0},
            /*10*/{0,0,0,0,0,0,1,0,0,1,0,1,0,0,1,0},
            /*11*/{0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,1},
            /*12*/{0,0,0,0,0,0,0,0,1,0,0,0,0,1,0,0},
            /*13*/{0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,0},
            /*14*/{0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1},
            /*15*/{0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0},
    };
    
    public int[][] getData() {
        return aja;
    }
    
    public int getNumOfShortestPahtes(int[][] edge) {
        int len = edge.length;
        int[] step = new int[len];
        int[] path = new int[len];
        path[0] = 1;
        Queue<Integer> q = new LinkedList<Integer>();
        q.add(0); //将起点放入
        
        while(!q.isEmpty()) {
            int element = q.remove();
            for(int j=1; j<len; j++) {
                if(edge[element][j] == 1) {
                    if(step[j] == 0 || step[j] > step[element] + 1) {
                        step[j] = step[element] + 1;
                        path[j] = path[element];
                        q.add(j);
                    }
                    else if(step[j] == step[element] + 1) 
                        path[j] += path[element];
                }
            }
        }
        return path[len-1];
    }
    
}

 

来源：https://www.cnblogs.com/little-YTMM/p/5448652.html