依照Vi的順序一一尋找每一組Vi的最短路徑ppt課件

1、GraphMSTMSTvPrims Algorithmv按照Vi的順序一一尋找每一組Vi的最短路徑，畫出spanning treevKruskal Algorithmv尋找整個圖中的最短路徑，在一一將其連接起來Prims AlgorithmvPrims algorithm has the property that the edges in the set A always form a single tree. vWe begin with some vertex v in a given graph G =(V, E)vDefining the initial set of vertice

2、s A. vThen, in each iteration, we choose a minimum-weight edge (u, v), connecting a vertex v in the set A to the vertex u outside of set A. Then vertex u is brought in to A. This process is repeated until a spanning tree is formed.Prims Algorithm (cont.)vPrim-MST(G) vSelect an arbitrary vertex to st

3、art the tree from. vWhile (there are still non-tree vertices) vSelect the edge of minimum weight between a tree and non-tree vertex vAdd the selected edge and vertex to the tree . vAnalysis: O(n2)Prims Algorithm (cont.)Total cost: 66練習Kruskals AlgorithmvThe Kruskal Algorithm starts with a forest whi

4、ch consists of n trees.vEach and everyone tree,consists only by one node and nothing else.vIn every step of the algorithm,two different trees of this forest are connected to a bigger tree.vTherefore ,we keep having less and bigger trees in our forest until we end up in a tree which is the minimum sp

5、anning treeKruskals Algorithm (cont.)The forest is constructed - with each node in a separate tree. The edges are placed in a priority queue. Until weve added n-1 edges, Extract the cheapest edge from the queue, If it forms a cycle, reject it, Else add it to the forest. Adding it to the forest will

6、join two trees together. Every step will have joined two trees in the forest together, so that at the end, there will only be one tree in T. Kruskals Algorithm (cont.)Total cost: 49練習Shortest Pathv在一個有向圖形G=(V,E)，G中每一個邊都有一個比例常數W(Weight)與之對應，假想象求G圖形中某一個頂VO到其它頂點的最少W總和之值，這類問題就稱為最短路徑問題(The Shortest Path

7、Problem)v單點對全部頂點的最短距離及一切頂點兩兩之間的最短距離。Dijkstras Algorithmv一個頂點到多個頂點通常運用Dijkstra演算法求得，Dijkstra的演算法如下：v假設S=Vi|ViV，且Vi在已發(fā)現的最短路徑，其中V0 S是起點。v假設w S，定義Dist(w)是從V0到w的最短路徑，這條路徑除了w外必屬於S。且有以下幾點特性：v假設u是目前所找到最短路徑之下一個節(jié)點，則u必屬於V-S集合中最小花費本錢的邊。Dijkstras Algorithm (cont.)假設u被選中，將u參與S集合中，則會產生目前的由V0到u最短路徑，對於w S ，DIST(w)被改

8、變成DIST(w)MinDIST(w),DIST(u)+COST(u,w)Dijkstra的方法只能求出某一點到其他頂點的最短距離，假設要求出圖形中任兩點甚至一切頂點間最短的距離，就必須運用Floyd演算法。Dijkstras Algorithm (cont.)vDijkstras algorithm can be expressed formally as follows: vG - arbitrary connected graph v0 - is the initial beginning vertex V - is the set of all vertices in the grap

9、h G S - set of all vertices with permanent labels n - number of vertices in G D - set of distances to v0 C - set of edges in G Dijkstras Algorithm (cont.)vDijkstra Algorithm (graph G, vertex v0) S=v0 For i = 1 to n Di = Cv0,i v For i = 1 to n-1 Choose a vertex w in V-S such that Dw is minimum Add w

10、to S For each vertex v in V-S Dv = min(Dv, Dw + Cw,v) 練習Floyd AlgorithmvFloyd演算法定義：vAkij=minAk-1ij,Ak-1ik+Ak-1 kj，1i，j nv其中Akij為從i到j不以註標大於k之頂點為間接路徑頂點時之最短路徑的花費。vA0ij=COSTij即A0便等於COSTvA0為頂點i到j間的直通距離。vAni,j代表i到j之最短路距離，即An便是我們所要求的最短路徑本錢矩陣。Floyd Algorithm (cont.)Example W = D0 =40520-30123123000000000123

11、123P =1235-324 D1 =405207-30123123000001000123123P =1235-324k = 1Vertex 1 can be intermediate node D12,3 = min( D02,3, D02,1+D01,3 )= min (, 7) = 7D13,2 = min( D03,2, D03,1+D01,2 )= min (-3,) = -340520-30123123D0 = D2 =405207-1-30123123000001200123123P =D21,3 = min( D11,3, D11,2+D12,3 )= min (5, 4+7

12、) = 5D23,1 = min( D13,1, D13,2+D12,1 )= min (, -3+2) = -11235-324 D1 =405207-30123123k = 2Vertices 1, 2 can be intermediate D3 =205207-1-30123123300001200123123P =D31,2 = min(D21,2, D21,3+D23,2 )= min (4, 5+(-3) = 2D32,1 = min(D22,1, D22,3+D23,1 )= min (2, 7+ (-1) = 2 D2 =405207-1-301231231235-324k

13、= 3Vertices 1, 2, 3 can be intermediate練習123414352Transitive Closure (cont.)vGiven a digraph G, the transitive closure of G is the digraph G* such thatvG* has the same vertices as Gvif G has a directed path from u to v (u v), G* has a directed edge from u to vvThe transitive closure provides reachab

14、ility information about a digraphBADCEBADCEGG*Transitive Closure (cont.)v建立一個路徑矩陣P，從這個矩陣中可以輕易的判斷圖形G中的任一兩點I跟j能否存在一路徑v先求得A1An An(aij)表示圖形node i 跟j存在為n的路徑總數v將矩陣A1.An全部相加，將總和存放在Bv將B矩陣中一切大於0的設定為1，將元素設定為0根1的矩陣，此矩陣的圖形G就是個長度為n的Transtive closureTransitive Closure (cont.)Warshall (int N, int A, intP) int i,j,

15、k; for (i = 0; i N; i+) for (j = 0; j N; j+) /* Theres a path if theres an edge */ Pij = Aij; for (k = 0; k N; k+) for (i = 0; i N; i+) for (j = 0; j N; j+) if (! Pij) Pij = Pik & Pkj; /* Warshall */AOV網路根本定義v以圖形嚴謹的定義來說，AOV網路就是一種有向圖形，在這個有向圖形中的每一個節(jié)點代表一項任務或必須執(zhí)行的動作，而那些有方向性的邊則代表了任務與任務之間存在的先後關係順序。也就是

16、說，表示必須處理先完Vi的任務，才可以進行Vj的任務。拓樸排序v拓樸排序的功能是將部分的排序partial ordering的關係轉換為線性的次序v具有這種特性的線性次序稱為拓樸序列topologcal orderv拓樸序列需是個acyclic graphv產生的拓樸序列必非獨一拓樸排序的步驟：v步驟1v尋找圖形中任何一個沒有先行者 predecessor 的點。v步驟2v輸出此頂點，並將此頂點的一切邊刪除。v步驟3v重複上兩個步驟處理一切頂點。C CD DE EG GH HF FI IK KJ JB BA A範例v先選取沒有predecessor的A，將其移除並宜出其相關的edgeC CD

17、DE EG GH HF FI IK KJ JB BTopological order:A BEG CFH DI JKAOE網路和臨界路徑(critical path)v臨界路徑就是AOE有向圖形從源頭頂點到目的頂點間所需花費時間最長的一條有方向性的路徑，當有一條以上的花費時間相等，而且都是最長，則這些路徑都稱為此AOE有向圖形的臨界路徑。相關練習v1048 106 miles to Chicago v1068 Islands and Bridges v1153 The Bottom of a Graph x*u$qZnVkShPdMaJ7F4C0z)w&s!pYmUjRfOcL9H6E

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