单源最短路径问题（分支限界法）

问题描述：在所给的有向图G中，每一边都有一个非负边权，要求图G的从源顶点s到目标顶点t之间的最短路径；

算法思想：使用优先队列式分支限界法

代码：

/************************************************************************//* 单源点最短路劲/* 优先队列式分支限界法 *//* author:yel_hb/************************************************************************/#include <iostream>using namespace std;#define MAX 9999//定义无穷大/***Graph类，用以存放有关图的所有信息*/class Graph {public://---------------------------//param int 初始节点编号//--------------------------void ShorestPaths(int);void ShowDist();Graph();private:int n;//图的节点个数int *prev;//存放顶点的前驱节点int **c;//存放图的邻接矩阵int *dist;//存放源点到各个顶点的距离};/***节点*/class MinHeapNode{friend Graph;public:int getI() {return i;}void setI(int ii){i = ii;}int getLength(){return length;}void setLength(int len){length = len;}private:int i;//顶点编号int length;//当前路长};/***最小堆*/class MinHeap{friend Graph;public:MinHeap();MinHeap(int);void DeleteMin(MinHeapNode &);void Insert(MinHeapNode);bool OutOfBounds();private:int length;MinHeapNode *node;};Graph::Graph(){int wi = 0; int yi = 0;cout<<"请输入图的节点个数：";cin>>n;cout<<"请输入图的邻接矩阵：(无穷大请以9999代替)" << endl;c = new int*[n+1];dist = new int[n+1];prev = new int[n+1];//------------------------------//初始化邻接矩阵//------------------------------for (wi = 0; wi <= n; wi++){c[wi] = new int[n+1];if (wi == 0){for (yi = 0; yi <= n; yi++){c[wi][yi] = 0;}}else{for (yi = 0; yi <= n; yi++){if (yi == 0){c[wi][yi] = 0;}else{cin >> c[wi][yi];}}}}//----------------------------------//初始化数组//----------------------------------for (wi = 0; wi <= n; wi++){dist[wi] = MAX;prev[wi] = 0;}}void Graph::ShowDist(){cout << "从源点到该节点的最短路径:" << endl;int i = 0;int temp = 0;for (i = 1; i <= n; i++){cout << "dist[" << i << "] = " << dist[i] << endl;}cout << "从源点到终点的最短路径长度为:" << dist[n] << endl;cout << "其路径为：";temp = n;while(temp != 0){if (prev[temp] == 0){cout << temp;}else{cout << temp << "->";}temp = prev[temp];}cout << endl;}void Graph::ShorestPaths(int v){MinHeap H(n); //最小堆MinHeapNode E; //扩展节点E.i = v;E.length = 0;dist[v] = 0;//搜索问题的解空间树while (true){int j = 0;for (j = 1; j <= n; j++){cout<<"c["<<E.i<<"]["<<j<<"]="<<c[E.i][j]<<endl;if ((c[E.i][j] != MAX) && (c[E.i][j] != 0)){//节点控制关系if (E.length + c[E.i][j] < dist[j]){dist[j] = E.length + c[E.i][j];prev[j] = E.i;//加入活结点优先队列//若节点为叶子节点，则不加入活结点队列if (j != n){MinHeapNode N;N.i = j;N.length = dist[j];H.Insert(N);}} else{H.DeleteMin(E);}}}if (H.OutOfBounds()){break;}cout<<"上一个扩展节点"<<E.i<<" "<<E.length<<endl;H.DeleteMin(E);cout<<"下一个扩展节点"<<E.i<<" "<<E.length<<endl;}}MinHeap::MinHeap(){length = 10;node = new MinHeapNode[length+1];for (int i = 0; i <= length; i++){node[i].setI(0);node[i].setLength(0);}}MinHeap::MinHeap(int n){length = n;node = new MinHeapNode[length+1];for (int i = 0; i <= length; i++){node[i].setI(0);node[i].setLength(0);}}/***取下一个扩展结点，并删除此节点**算法实现其实是用下一个节点的信息替代现有节点的数据**首先在现有的节点中，找出length最短的节点**然后将此节点的数据替换原有的数据*/void MinHeap::DeleteMin(MinHeapNode &E){int i = 0;int j = 0;j = E.getI();//用来删除原来的扩展节点node[j].setI(0);//置零node[j].setLength(0);//置零int temp = MAX;//-------------------------------------//选择可扩展节点中length域最小的可扩展节点//将所选择的扩展节点的数值替换原有的扩展节//点的值，最后在可扩展节点队列中删除原扩展//节点，删除方式为：所有域置零//-------------------------------------for (i = 1; i <= length; i++){if ((node[i].getLength() < temp) && (node[i].getLength() != 0)){E.setI(i);E.setLength(node[i].getLength());temp = node[i].getLength();//temp中始终为最小值}}}/***加入最小堆**此处添加按节点编号添加，即对应的节点编号添加时**对应队列中相应的编号，即节点5则添加到队列中5号**位置*/void MinHeap::Insert(MinHeapNode N){node[N.getI()].setI(N.getI());node[N.getI()].setLength(N.getLength());}/***判断最小堆是否为空*/bool MinHeap::OutOfBounds(){int i = 0;bool flag = true;for (i = 1; i <= length; i++){if (node[i].getI() != 0){flag = false;}}return flag;}int main(){Graph graph;graph.ShorestPaths(1);graph.ShowDist();return 0;}