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  • Queue 应用——拓扑排序

    1. 拓扑排序

    题目描述:对一个有向无环图(Directed Acyclic Graph, DAG)G进行拓扑排序,是将G中所有顶点排成线性序列,是的图中任意一堆顶点u和v,若边(u, v)在E(G)中,则u在线性序列中出现在v之前。

    如:

    分析:

    1)首先我们要将图G存入一个邻接矩阵中,保存该图;

    2)计算每个顶点的入度,存储一个一维数组中;

    3)从有向图中选择一个没有前驱(即入度为0)的节点并输出;

    4)从图中删除该节点,并且删除从该节点发出的全部有向边;

    5)重复上面两个步骤,直至剩余的图中不再存在没有前驱的节点为止。

    进一步思考:

    1)拓扑排序的本质是不断输出入度为0的点,这种方法可以用于判断图中是否有环;

    2)拓扑排序其实给出的是节点之间的偏序关系; 

    Answer:

    class TopologySort {
        private int[][] aja = {{0,1,0,0,0,1,1,0,0,0,0,0,0},
                     {0,0,0,0,0,0,0,0,0,0,0,0,0},
                  {1,0,0,1,0,0,0,0,0,0,0,0,0},
                  {0,0,0,0,0,1,0,0,0,0,0,0,0},
                  {0,0,0,0,0,0,0,0,0,0,0,0,0},
                  {0,0,0,0,1,0,0,0,0,0,0,0,0},
                  {0,0,0,0,1,0,0,0,0,1,0,0,0},
                  {0,0,0,0,0,0,1,0,0,0,0,0,0},
                  {0,0,0,0,0,0,0,1,0,0,0,0,0},
                  {0,0,0,0,0,0,0,0,0,0,1,1,1},
                  {0,0,0,0,0,0,0,0,0,0,0,0,0},
                  {0,0,0,0,0,0,0,0,0,0,0,0,1},
                  {0,0,0,0,0,0,0,0,0,0,0,0,0}};
        
        public int[][] getEdge() {
            return aja;
        }
        
        
        /**
         * 该函数根据邻接矩阵计算每个节点的入度
         * @param edge
         * @return
         */
        public int[] getInDegree(int[][] edge) {
            int len = edge.length;
            int[] inDegree = new int[len];
            for(int j=0; j<len; j++) {
                int count = 0;
                for(int i=0; i<len; i++) 
                    if(edge[i][j] == 1)
                        count++;
                inDegree[j] = count;
            }
            return inDegree;
        }
                  
        public List<Integer> topoSort(int[][] edge) {
            List<Integer> res = new ArrayList<Integer>();
            int len = edge.length;
            int[] inDegree = getInDegree(edge);
            
            Queue<Integer> q = new LinkedList<Integer>();
            //将入度为0的节点压入队列中
            for(int i=0; i<inDegree.length; i++) 
                if(inDegree[i] == 0) {
                    q.add(i);
                    res.add(i);
                }
            
            //对于队列中的元素(入度为0的元素)
            while(!q.isEmpty()) {
                int element = q.remove();
                for(int j=0; j<len; j++) {
                    if(edge[element][j] == 1)  {
                        inDegree[j]--;
                        if(inDegree[j] == 0) {
                            q.add(j);
                            res.add(j);
                        }
                    }
                }
            }
            return res;
        }
    }
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  • 原文地址:https://www.cnblogs.com/little-YTMM/p/5448442.html
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