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; }