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,存储一个以为数组中;
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又想边;
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; } }
时间: 2024-11-05 15:54:09