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(huà)出spanning treevKruskal Algorithmv尋找整個(gè)圖中的最短路徑，在一一將其連接起來(lái)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(xí)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(xí)Shortest Pathv在一個(gè)有向圖形G=(V,E)，G中每一個(gè)邊都有一個(gè)比例常數(shù)W(Weight)與之對(duì)應(yīng)，假想象求G圖形中某一個(gè)頂VO到其它頂點(diǎn)的最少W總和之值，這類(lèi)問(wèn)題就稱(chēng)為最短路徑問(wèn)題(The Shortest Path

7、Problem)v單點(diǎn)對(duì)全部頂點(diǎn)的最短距離及一切頂點(diǎn)兩兩之間的最短距離。Dijkstras Algorithmv一個(gè)頂點(diǎn)到多個(gè)頂點(diǎn)通常運(yùn)用Dijkstra演算法求得，Dijkstra的演算法如下：v假設(shè)S=Vi|ViV，且Vi在已發(fā)現(xiàn)的最短路徑，其中V0 S是起點(diǎn)。v假設(shè)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(diǎn)特性：v假設(shè)u是目前所找到最短路徑之下一個(gè)節(jié)點(diǎn)，則u必屬於V-S集合中最小花費(fèi)本錢(qián)的邊。Dijkstras Algorithm (cont.)假設(shè)u被選中，將u參與S集合中，則會(huì)產(chǎn)生目前的由V0到u最短路徑，對(duì)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(diǎn)到其他頂點(diǎn)的最短距離，假設(shè)要求出圖形中任兩點(diǎn)甚至一切頂點(diǎn)間最短的距離，就必須運(yùn)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(xí)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(luò)不以註標(biāo)大於k之頂點(diǎn)為間接路徑頂點(diǎn)時(shí)之最短路徑的花費(fèi)。vA0ij=COSTij即A0便等於COSTvA0為頂點(diǎn)i到j(luò)間的直通距離。vAni,j代表i到j(luò)之最短路距離，即An便是我們所要求的最短路徑本錢(qián)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(xí)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(gè)路徑矩陣P，從這個(gè)矩陣中可以輕易的判斷圖形G中的任一兩點(diǎn)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(shù)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(shè)定為1，將元素設(shè)定為0根1的矩陣，此矩陣的圖形G就是個(gè)長(zhǎng)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(wǎng)路根本定義v以圖形嚴(yán)謹(jǐn)?shù)亩x來(lái)說(shuō)，AOV網(wǎng)路就是一種有向圖形，在這個(gè)有向圖形中的每一個(gè)節(jié)點(diǎn)代表一項(xiàng)任務(wù)或必須執(zhí)行的動(dòng)作，而那些有方向性的邊則代表了任務(wù)與任務(wù)之間存在的先後關(guān)係順序。也就是

16、說(shuō)，表示必須處理先完Vi的任務(wù)，才可以進(jìn)行Vj的任務(wù)。拓樸排序v拓樸排序的功能是將部分的排序partial ordering的關(guān)係轉(zhuǎn)換為線(xiàn)性的次序v具有這種特性的線(xiàn)性次序稱(chēng)為拓樸序列topologcal orderv拓樸序列需是個(gè)acyclic graphv產(chǎn)生的拓樸序列必非獨(dú)一拓樸排序的步驟：v步驟1v尋找圖形中任何一個(gè)沒(méi)有先行者 predecessor 的點(diǎn)。v步驟2v輸出此頂點(diǎn)，並將此頂點(diǎn)的一切邊刪除。v步驟3v重複上兩個(gè)步驟處理一切頂點(diǎn)。C CD DE EG GH HF FI IK KJ JB BA A範(fàn)例v先選取沒(méi)有predecessor的A，將其移除並宜出其相關(guān)的edgeC CD

17、DE EG GH HF FI IK KJ JB BTopological order:A BEG CFH DI JKAOE網(wǎng)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(diǎn)到目的頂點(diǎn)間所需花費(fèi)時(shí)間最長(zhǎng)的一條有方向性的路徑，當(dāng)有一條以上的花費(fèi)時(shí)間相等，而且都是最長(zhǎng)，則這些路徑都稱(chēng)為此AOE有向圖形的臨界路徑。相關(guān)練習(xí)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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