Algorithms Fourth Edition
Written By Robert Sedgewick & Kevin Wayne
Translated By 謝路雲
Chapter 4 Section 4 最短路徑算法
圖是強連通的數組
權重都爲正數據結構
最短路徑不必定是惟一的,咱們只找出其中一條性能
可能存在平行邊和自環(但咱們會忽略自環)this
有向邊,因此新增方法from() 和 to()spa
public class DirectedEdge { private final int v; // edge source private final int w; // edge target private final double weight; // edge weight public DirectedEdge(int v, int w, double weight) { this.v = v; this.w = w; this.weight = weight; } public double weight() { return weight; } public int from() { return v; } public int to() { return w; } public String toString() { return String.format("%d->%d %.2f", v, w, weight); } }
public class EdgeWeightedDigraph { private final int V; // number of vertices private int E; // number of edges private Bag<DirectedEdge>[] adj; // adjacency lists public EdgeWeightedDigraph(int V) { this.V = V; this.E = 0; adj = (Bag<DirectedEdge>[]) new Bag[V]; for (int v = 0; v < V; v++) adj[v] = new Bag<DirectedEdge>(); } public EdgeWeightedDigraph(In in)// See Exercise 4.4.2. public int V() { return V; } public int E() { return E; } public void addEdge(DirectedEdge e) { adj[e.from()].add(e); E++; } public Iterable<Edge> adj(int v) { return adj[v]; } public Iterable<DirectedEdge> edges() { Bag<DirectedEdge> bag = new Bag<DirectedEdge>(); for (int v = 0; v < V; v++) for (DirectedEdge e : adj[v]) bag.add(e); } }
兩條路徑code
s --> worm
s --> v , v -> w隊列
比較哪一條路徑更短,記錄更短的那個邊。圖片
若 路徑1 < 路徑2,原路徑 s --> w 已經最短,不更新。
邊 v -> w 失效
若 路徑1 > 路徑2,新路徑 s --> v , v -> w 更短,更新,放鬆成功
路徑 s --> w 中 原指向w的那一條邊失效
private void relax(DirectedEdge e) { int v = e.from(), w = e.to(); if (distTo[w] > distTo[v] + e.weight()) { distTo[w] = distTo[v] + e.weight(); edgeTo[w] = e;//記錄的是邊,而不是點 } }
private void relax(EdgeWeightedDigraph G, int v) { for (DirectedEdge e : G.adj(v)) { int w = e.to(); if (distTo[w] > distTo[v] + e.weight()) { distTo[w] = distTo[v] + e.weight(); edgeTo[w] = e; } } }
當且僅當 v -> w 的任意一條邊e都知足 distTo[w] <= distTo[v] + e.weight(),它們是最短路徑
distTo[s]=0, distTo[v]=INFINITY(v≠s)
放鬆G中的任意邊,直到不存在有效邊爲止
非負權重
distTo[s]=0, distTo[v]=INFINITY(v≠s)
將distTo[]中 離頂點s最近的非樹頂點 放鬆, 並加入到樹中
重複2,直到全部頂點都在樹中 或者 全部的非樹頂點的distTo[]值均爲無窮大
複雜度
空間:V
時間:ElogV
public class DijkstraSP { private DirectedEdge[] edgeTo; //記錄路徑 private double[] distTo; //記錄權重 private IndexMinPQ<Double> pq; //優先隊列 public DijkstraSP(EdgeWeightedDigraph G, int s) { edgeTo = new DirectedEdge[G.V()]; distTo = new double[G.V()]; pq = new IndexMinPQ<Double>(G.V()); for (int v = 0; v < G.V(); v++) distTo[v] = Double.POSITIVE_INFINITY; //初始化距離爲正無窮 distTo[s] = 0.0; //頂點s到頂點s的距離固然爲0 pq.insert(s, 0.0); //第一次遇到頂點,插入 while (!pq.isEmpty()) //直到全部頂點都失效(即全部頂點都已加入到最短路徑中) relax(G, pq.delMin()); //鬆弛,每次鬆弛,都從隊列中刪除一個點(即加入到最短路徑中) } private void relax(EdgeWeightedDigraph G, int v) { for (DirectedEdge e : G.adj(v)) { //遍歷從v出發的每一條邊 int w = e.to(); // v -> w if (distTo[w] > distTo[v] + e.weight()) { //若是存在比目前s-->w更短的路徑, s-->v,v->w distTo[w] = distTo[v] + e.weight(); //更新距離/權重 edgeTo[w] = e; //更新路徑 if (pq.contains(w)) // 隊列中有這個點 pq.change(w, distTo[w]); //更新,更新隊列中w的權重distTo[w]的值 else //隊列中沒有這個點 pq.insert(w, distTo[w]); //插入,把點w和權重distTo[w]做爲總體插入到隊列中 } } } public double distTo(int v) { return distTo[v]; } public boolean hasPathTo(int v) { return distTo[v] < Double.POSITIVE_INFINITY; } public Iterable<DirectedEdge> pathTo(int v) { if (!hasPathTo(v)) return null; Stack<DirectedEdge> path = new Stack<DirectedEdge>(); for (DirectedEdge e = edgeTo[v]; e != null; e = edgeTo[e.from()]) path.push(e); return path; } }
DijkstraSP算法 VS Prim 算法
DijkstraSP算法 每次添加的都是離起點最近的非樹頂點
Prim 算法 每次添加的是離樹頂點最近的非樹頂點
不須要數組marked[],!marked[v] 等價於 distTo[v]無窮大
DijkstraSP算法忽略relax()方法中的distTo[v]部分的代碼,便可獲得Prim算法的即時版本
頂點s,v的最短路徑怎麼求?
用DijkstraSP算法,並在優先隊列中刪除頂點v後中止
任意頂點對的最短路徑怎麼求?
public class DijkstraAllPairsSP { private DijkstraSP[] all; DijkstraAllPairsSP(EdgeWeightedDigraph G) { all = new DijkstraSP[G.V()]; for (int v = 0; v < G.V(); v++) all[v] = new DijkstraSP(G, v); } Iterable<Edge> path(int s, int t) { return all[s].pathTo(t); } double dist(int s, int t) { return all[s].distTo(t); } }
更快更簡單更好的算法
線性時間
可以處理負權重
可以解決其餘相關問題,eg 距離最長
distTo[s]=0, distTo[v]=INFINITY(v≠s)
按照 拓撲順序 放鬆全部頂點
複雜度
時間: E+V
空間: V
public class AcyclicSP { private DirectedEdge[] edgeTo; private double[] distTo; public AcyclicSP(EdgeWeightedDigraph G, int s) { edgeTo = new DirectedEdge[G.V()]; distTo = new double[G.V()]; for (int v = 0; v < G.V(); v++) distTo[v] = Double.POSITIVE_INFINITY; distTo[s] = 0.0; Topological top = new Topological(G); //只增長了這一個!!!就把性能提升了!!! for (int v : top.order()) relax(G, v); } private void relax(EdgeWeightedDigraph G, int v) { for (DirectedEdge e : G.adj(v)) { //遍歷從v出發的每一條邊 int w = e.to(); // v -> w if (distTo[w] > distTo[v] + e.weight()) { //若是存在比目前s-->w更短的路徑, s-->v,v->w distTo[w] = distTo[v] + e.weight(); //更新距離/權重 edgeTo[w] = e; //更新路徑 } } } public double distTo(int v) { return distTo[v]; } public boolean hasPathTo(int v) { return distTo[v] > Double.NEGATIVE_INFINITY; } public Iterable<DirectedEdge> pathTo(int v) { if (!hasPathTo(v)) return null; Stack<DirectedEdge> path = new Stack<DirectedEdge>(); for (DirectedEdge e = edgeTo[v]; e != null; e = edgeTo[e.from()]) { path.push(e); } return path; } }
作一個副本,將無環加權有向圖的權重取反便可。(相關操做就是判斷的不等號符號改反,初始值設爲負無窮)
副本的最短路徑即爲原圖的最長路徑。
public class AcyclicLP { private double[] distTo; // distTo[v] = distance of longest s->v path private DirectedEdge[] edgeTo; // edgeTo[v] = last edge on longest s->v path public AcyclicLP(EdgeWeightedDigraph G, int s) { distTo = new double[G.V()]; edgeTo = new DirectedEdge[G.V()]; for (int v = 0; v < G.V(); v++) distTo[v] = Double.NEGATIVE_INFINITY; distTo[s] = 0.0; // relax vertices in toplogical order Topological topological = new Topological(G); if (!topological.hasOrder()) throw new IllegalArgumentException("Digraph is not acyclic."); for (int v : topological.order()) { for (DirectedEdge e : G.adj(v)) relax(e); } } // relax edge e, but update if you find a *longer* path private void relax(DirectedEdge e) { int v = e.from(), w = e.to(); if (distTo[w] < distTo[v] + e.weight()) { distTo[w] = distTo[v] + e.weight(); edgeTo[w] = e; } } public double distTo(int v) { return distTo[v]; } public boolean hasPathTo(int v) { return distTo[v] > Double.NEGATIVE_INFINITY; } public Iterable<DirectedEdge> pathTo(int v) { if (!hasPathTo(v)) return null; Stack<DirectedEdge> path = new Stack<DirectedEdge>(); for (DirectedEdge e = edgeTo[v]; e != null; e = edgeTo[e.from()]) { path.push(e); } return path; } }
爲每個點添加一個點做爲任務的結束點,並從結束點出髮指向充分條件。
添加一個起點,一個終點。
輸入文本格式(解讀,共10個任務,任務0耗時41秒,需在1,7,9以前完成...)
10 41.0 1 7 9 51.0 2 50.0 36.0 38.0 45.0 21.0 3 8 32.0 3 8 32.0 2 29.0 4 6
public class CPM { public static void main(String[] args) { int N = StdIn.readInt(); StdIn.readLine(); EdgeWeightedDigraph G; G = new EdgeWeightedDigraph(2 * N + 2); int s = 2 * N, t = 2 * N + 1; //s爲起點,t爲終點 for (int i = 0; i < N; i++) { String[] a = StdIn.readLine().split("\\s+"); double duration = Double.parseDouble(a[0]); G.addEdge(new DirectedEdge(i, i + N, duration)); //添加一個點做爲任務的結束點 G.addEdge(new DirectedEdge(s, i, 0.0)); //和起點相連 G.addEdge(new DirectedEdge(i + N, t, 0.0));//任務的結束點和終點相連 for (int j = 1; j < a.length; j++) { int successor = Integer.parseInt(a[j]);//讀取充分條件 G.addEdge(new DirectedEdge(i + N, successor, 0.0));//任務的結束點和充分條件相連 } } AcyclicLP lp = new AcyclicLP(G, s); //最長路徑 StdOut.println("Start times:"); for (int i = 0; i < N; i++) StdOut.printf("%4d: %5.1f\n", i, lp.distTo(i)); StdOut.printf("Finish time: %5.1f\n", lp.distTo(t)); } }
再加一個限制:deadline,也就是截止時間限制(相對某個任務的截止時間,好比2號任務必須在4號任務啓動的12個單位時間內開始)。
方法是同上面同樣構造圖,同時會添加負權重邊,再將全部邊取反,而後求最短路徑
最短路徑存在則可行(沒有負權重環就是可行的調度)。
考慮有環也可能負邊的最短路徑問題
負權重環會致使繞圈現象,所以負權重環存在求不出最短路徑
以任意順序放鬆全部邊
重複V輪
複雜度
時間: EV
空間: V
public BellmanFord_BruceAlg() { for (int pass = 0; pass < G.V(); pass++) //第i輪 for (v = 0; v < G.V(); v++) //在每一輪中放鬆全部邊 for (DirectedEdge e : G.adj(v)) relax(e); }
在後幾輪中,不少邊的放鬆都不會成功
只有上一輪distTo[]的值發生改變的頂點指出的邊,才能改變其餘頂點的distTo[]值
用隊列記錄這樣的頂點
public class BellmanFordSP { private double[] distTo; // length of path to v private DirectedEdge[] edgeTo; // last edge on path to v private boolean[] onQ; // Is this vertex on the queue? private Queue<Integer> queue; // vertices being relaxed private int cost; // number of calls to relax() private Iterable<DirectedEdge> cycle; // negative cycle in edgeTo[]? public BellmanFordSP(EdgeWeightedDigraph G, int s) { distTo = new double[G.V()]; edgeTo = new DirectedEdge[G.V()]; onQ = new boolean[G.V()]; queue = new Queue<Integer>(); for (int v = 0; v < G.V(); v++) distTo[v] = Double.POSITIVE_INFINITY; distTo[s] = 0.0; queue.enqueue(s); onQ[s] = true; while (!queue.isEmpty() && !this.hasNegativeCycle()) { int v = queue.dequeue(); onQ[v] = false; relax(v); } } private void relax(EdgeWeightedDigraph G, int v){ for (DirectedEdge e : G.adj(v){ int w = e.to(); if (distTo[w] > distTo[v] + e.weight()){ distTo[w] = distTo[v] + e.weight(); edgeTo[w] = e; if (!onQ[w]){ q.enqueue(w); onQ[w] = true; } } if (cost++ % G.V() == 0) //竟然在這裏判斷是否循環了V輪。。。 findNegativeCycle(); } } public double distTo(int v) // standard client query methods public boolean hasPathTo(int v) // for SPT implementatations public Iterable<Edge> pathTo(int v) // (See page 649.) private void findNegativeCycle() { int V = edgeTo.length; EdgeWeightedDigraph spt; spt = new EdgeWeightedDigraph(V); for (int v = 0; v < V; v++) if (edgeTo[v] != null) spt.addEdge(edgeTo[v]); EdgeWeightedCycleFinder cf; cf = new EdgeWeightedCycleFinder(spt); cycle = cf.cycle(); } public boolean hasNegativeCycle() { return cycle != null; } public Iterable<Edge> negativeCycle() { return cycle; } }