一.最小生成树
永远遵循取最小权值的原则
如一个顶点到多个顶点,取最小权值
1.添加边结构
var Edge = function (sv, dv, d) { //起始点 this.srcVert = sv; //重点 this.destVert = dv; //权值 this.distance = d; }
2.添加优先级队列
var PriorityQ = function () { this.queArray = []; this.Max_Size = 10; this.length = 0; }; PriorityQ.prototype.Insert = function (item) { var i; //compute distance index for (i = 0; i < this.length; i++) { if (item.distance >= this.queArray[i].distance) break; } //move for (var j = this.length - 1; j >= i; j--) { this.queArray[j + 1] = this.queArray[j]; } this.queArray[i] = item; this.length++; }; PriorityQ.prototype.remove = function () { if (arguments.length == 0) return this.queArray[--this.length]; if (arguments.length == 1) { for (var j = arguments[0]; j < this.length - 1; j++) { this.queArray[j] = this.queArray[j + 1]; } this.length--; } }; PriorityQ.prototype.peek = function () { if (arguments.length == 0) return this.queArray[this.length - 1]; if (arguments.length == 1) { return this.queArray[arguments[0]]; } }; PriorityQ.prototype.IsEmpty = function () { return this.length == 0; } PriorityQ.prototype.find = function (dex) { for (var j = 0; j < this.length; j++) { if (this.queArray[j].destVert == dex) return j; } return -1; }3.1 标记当前项
this.getCurrentVertex().isInTree = true; nTree++;
3.2将符合条件的权值放入队列中
for (var j = 0; j < this.length; j++) { //skip //if it's us //in the tree //no edge if (j == this.currentVertIndex) continue; if (this.vertexList[j].isInTree) continue; var distance = this.adjMat[this.currentVertIndex][j]; if (distance == 0) continue; //put in pq //ensure index and distance this.putInPQ(j, distance); }如遇到起点相同的,则比较distance,放弃distance长的那个.
Graph.prototype.putInPQ=function(newVert, newDist) { var queueIndex = this.thePQ.find(newVert); if (queueIndex != -1) { var tempEdge = this.thePQ.peek(queueIndex); var oldDist = tempEdge.distance; //compare distance if (oldDist > newDist) { this.thePQ.remove(queueIndex); var theEdge = new Edge(this.currentVertIndex, newVert, newDist); this.thePQ.Insert(theEdge); } } else { //insert directly var theEdge = new Edge(this.currentVertIndex, newVert, newDist); this.thePQ.Insert(theEdge); } }
3.4.取最小权值
if (this.thePQ.length == 0) return; //remove minium distance from dq var theEdge = this.thePQ.remove(); var srcVert = theEdge.srcVert; this.currentVertIndex = theEdge.destVert; //display this.vertexList[srcVert].Display(); this.vertexList[this.currentVertIndex].Display();完整示例
Graph.prototype.getCurrentVertex=function() { return this.vertexList[this.currentVertIndex]; } Graph.prototype.mstw = function () { var nTree = 0; while (nTree < this.length - 1) { //put CurrentVertex in tree this.getCurrentVertex().isInTree = true; nTree++; for (var j = 0; j < this.length; j++) { //skip //if it's us //in the tree //no edge if (j == this.currentVertIndex) continue; if (this.vertexList[j].isInTree) continue; var distance = this.adjMat[this.currentVertIndex][j]; if (distance == 0) continue; //put in pq //ensure index and distance this.putInPQ(j, distance); } if (this.thePQ.length == 0) return; //remove minium distance from dq var theEdge = this.thePQ.remove(); var srcVert = theEdge.srcVert; this.currentVertIndex = theEdge.destVert; //display this.vertexList[srcVert].Display(); this.vertexList[this.currentVertIndex].Display(); } } Graph.prototype.putInPQ=function(newVert, newDist) { var queueIndex = this.thePQ.find(newVert); if (queueIndex != -1) { var tempEdge = this.thePQ.peek(queueIndex); var oldDist = tempEdge.distance; //compare distance if (oldDist > newDist) { this.thePQ.remove(queueIndex); var theEdge = new Edge(this.currentVertIndex, newVert, newDist); this.thePQ.Insert(theEdge); } } else { //insert directly var theEdge = new Edge(this.currentVertIndex, newVert, newDist); this.thePQ.Insert(theEdge); } }
二.求最短路径
迪杰斯特拉（Dijkstra）算法求顶点0到其他各顶点的最短路径
1.用于保存之前顶点到当前顶点的距离
class DistPar // distance and parent { // items stored in sPath array public int distance; // distance from start to this vertex public int parentVert; // current parent of this vertex // ------------------------------------------------------------- public DistPar(int pv, int d) // constructor { distance = d; parentVert = pv; } // ------------------------------------------------------------- } // end class DistPar
2.将第一个顶点标记为在树中
int startTree = 0; // start at vertex 0 vertexList[startTree].isInTree = true; nTree = 1; // put it in tree
3.初始化读取顶点到各个顶点的距离,如果没有边则距离为无穷大
// transfer row of distances from adjMat to sPath for(int j=0; j<nVerts; j++) { int tempDist = adjMat[startTree][j]; sPath[j] = new DistPar(startTree, tempDist); }
4.进入回圈
4.1在各点中取距离最小的点
int indexMin = getMin(); // get minimum from sPath int minDist = sPath[indexMin].distance;
4.2记录距离最小的点的索引值和距离,并把该顶点标记入树
if(minDist == INFINITY) // if all infinite { // or in tree, System.out.println("There are unreachable vertices"); break; // sPath is complete } else { // reset currentVert currentVert = indexMin; // to closest vert startToCurrent = sPath[indexMin].distance; // minimum distance from startTree is // to currentVert, and is startToCurrent } // put current vertex in tree vertexList[currentVert].isInTree = true;4.3 当把当前索引顶点记录之后,然后获取该顶点到各顶点的距离
int column = 1; // skip starting vertex while(column < nVerts) // go across columns { int currentToFringe = adjMat[currentVert][column]; }4.4记录上个距离和当前距离的和
int startToFringe = startToCurrent + currentToFringe;
4.5与第一个顶点到该节点的距离和startToFringe 进行比较,然后取短路径.并更新sPath的顶点和距离.
比如A到B是60,B到C是50,A到C是150,那么A经过B再到C的路径小于A直接到C的距离,则产生了最小路径
int sPathDist = sPath[column].distance; // compare distance from start with sPath entry if(startToFringe < sPathDist) // if shorter, { // update sPath sPath[column].parentVert = currentVert; sPath[column].distance = startToFringe; }完整（核心）
public void adjust_sPath() { // adjust values in shortest-path array sPath int column = 1; // skip starting vertex while(column < nVerts) // go across columns { // if this column's vertex already in tree, skip it if( vertexList[column].isInTree ) { column++; continue; } // calculate distance for one sPath entry // get edge from currentVert to column int currentToFringe = adjMat[currentVert][column]; // add distance from start int startToFringe = startToCurrent + currentToFringe; // get distance of current sPath entry int sPathDist = sPath[column].distance; // compare distance from start with sPath entry if(startToFringe < sPathDist) // if shorter, { // update sPath sPath[column].parentVert = currentVert; sPath[column].distance = startToFringe; } column++; } // end while(column < nVerts) }等sPath更新完毕后就是最小路径了
输出
Shortest paths
CurrentVert 1 distance 50
before begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 0 2 InTree false distance 1000000
ParentVert/SelfVert 0 3 InTree false distance 80
ParentVert/SelfVert 0 4 InTree false distance 1000000
before end sPath----------------
Start/Vert/column 0 1 2 distance 60 total undirect 110 sPathDist direct 1000000
2 changed
Start/Vert/column 0 1 3 distance 90 total undirect 140 sPathDist direct 80
Start/Vert/column 0 1 4 distance 1000000 total undirect 1000050 sPathDist direct 1000000
adjusted begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 1 2 InTree false distance 110
ParentVert/SelfVert 0 3 InTree false distance 80
ParentVert/SelfVert 0 4 InTree false distance 1000000
adjusted end sPath----------------
CurrentVert 3 distance 80
before begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 1 2 InTree false distance 110
ParentVert/SelfVert 0 3 InTree true distance 80
ParentVert/SelfVert 0 4 InTree false distance 1000000
before end sPath----------------
Start/Vert/column 0 3 2 distance 20 total undirect 100 sPathDist direct 110
2 changed
Start/Vert/column 0 3 4 distance 70 total undirect 150 sPathDist direct 1000000
4 changed
adjusted begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 3 2 InTree false distance 100
ParentVert/SelfVert 0 3 InTree true distance 80
ParentVert/SelfVert 3 4 InTree false distance 150
adjusted end sPath----------------
CurrentVert 2 distance 100
before begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 3 2 InTree true distance 100
ParentVert/SelfVert 0 3 InTree true distance 80
ParentVert/SelfVert 3 4 InTree false distance 150
before end sPath----------------
Start/Vert/column 0 2 4 distance 40 total undirect 140 sPathDist direct 150
4 changed
adjusted begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 3 2 InTree true distance 100
ParentVert/SelfVert 0 3 InTree true distance 80
ParentVert/SelfVert 2 4 InTree false distance 140
adjusted end sPath----------------
CurrentVert 4 distance 140
before begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 3 2 InTree true distance 100
ParentVert/SelfVert 0 3 InTree true distance 80
ParentVert/SelfVert 2 4 InTree true distance 140
before end sPath----------------
adjusted begin sPath----------------
ParentVert/SelfVert 0 0 InTree true distance 1000000
ParentVert/SelfVert 0 1 InTree true distance 50
ParentVert/SelfVert 3 2 InTree true distance 100
ParentVert/SelfVert 0 3 InTree true distance 80
ParentVert/SelfVert 2 4 InTree true distance 140
adjusted end sPath----------------
A=inf(A) B=50(A) C=100(D) D=80(A) E=140(C)
步骤总结:
1.选定一个顶点为根记做A
2.选择该顶点的最小边记做B,并记录这条边与A相交的顶点C,并标记入树C
3.计算C点到各顶点(不在树中的)的边的长度并与(A到各顶点的边)相比较,选择路径小者对A进行更新，并记录该顶点(以便排序顶点输出)
4.重复2-3步骤，等所有顶点都在树中时,则A到各顶点的最小路径更新完毕
这属于一个贪婪算法,总是先选最小的权值
终于搞懂了
理解后,其实也不难.多看看中间步骤,一步到位很难理解
三.弗洛伊德（Floyed）算法求每一对顶点之间的最短路径的