单源点最短路径

单源点最短路径的C实现

求解单源点最短路径的算法：

• 1构造所有节点间距离的矩阵（二维数组），无直接路径的为无穷
• 2构造到所有节点最短路径上前一结点数组nodeArray，初始全为源点
• 3构造源点到其他各结点的距离数组distanceArray，无直接路径的为无穷
• 4从数组distanceArray中选出距离最小的结点node,并入集合s（初始为空）
• 5遍历遍历每个不在s集合中的其他结点dot（distanceArray数组），若源点到node距离与node到dot的距离之和小于源点到dot距离，则更新distanceArray对应项的值，并同时更新nodeArray的值为node
• 6重复4-5直至无可选结点
• 7通过nodeArray从某结点出发不断寻找其前一结点，直至源点，此路径即为源点到该结点的最短路径

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具体实现

#include <iostream>#include <stdio.h>#include <stdlib.h>using namespace std;/**自定义比较大小，负数均视为无穷大，其他大小正常x<y,return true*/bool lessThen(float x, float y);/**长度为len的s数组是否包含e包含e，return true*/bool contain(int* s, int len, int e);/**得到到达e的最短路径字符串描述keys[i]的值为起点到i的前一个结点索引*/string getPath(int *keys, int e);/**以size*size的二维数组paths为前提寻找最短路径，返回有效长度为len一维数组keyskeys[i]的值为起点到i的前一个结点索引*/int* findShortestPath(float **paths, int size, int& len);int main(){    //节点数    int size = 0;    //读入节点数    cout<<"请输入节点数：";    cin>>size;    cout<<endl;    if(size <= 0)    {        cout<<"节点数非正数"<<endl;        system("pause");        return 0;    }    int i;    int j;    //路径集合size*size    float** paths = new float*[size];    for(i = 0; i < size; i++)    {        paths[i] = new float[size];    }    //读入路径集合    //readPaths(paths);    for(i = 0; i < size; i++)    {        for(j = 0; j < size; j++)        {            if(i == j)                paths[i][j] = 0;            else            {                cout<<"请输入节点"<<i<<"到节点"<<j<<"的距离(负数表示无穷大):";                cin>>paths[i][j];            }        }        cout<<endl;    }    //计算最短路径    int len;    int* keys = findShortestPath(paths, size, len);    string path;    for(i = 1; i < size; i++)    {        cout<<"到节点"<<i<<"的最短距离:";        if(paths[0][i] > 0)            cout<<paths[0][i];        else            cout<<"无穷大";        cout<<endl;        path = getPath(keys, i);        cout<<"到节点"<<i<<"的最短路径:"<<path<<endl<<endl;;    }    //释放内存    if(keys)        delete keys;    for(i = 0; i < size; i++)    {        if(paths[i])            delete paths[i];    }    if(paths)        delete paths;    system("pause");    return 0;}bool contain(int* s, int len, int e){    while(len > 0)    {        if(*s == e)            return true;        s++;        len--;    }    return false;}string getPath(int *keys, int e){    string path;    char* temp = new char[20];    while(1)    {        sprintf(temp, "%d", e);        path.append(temp);        if(e == 0)            break;        e = keys[e];        path.append(" <- ");    }    delete temp;    return path;}int* findShortestPath(float **paths, int size, int& len){    len = 0;    int* keys = new int[size];    int* shortest = new int[size];    shortest[len] = 0;    len++;    int index = 0;    int i;    int min;    int temp;    for(i = 0; i < size; i++)        keys[i] = 0;    while(len < size - 1)    {        //遍历找到距离最短的节点        min = -1;        for(i = 1; i < size; i++)        {            if(!contain(shortest, len, i) && lessThen(paths[0][i], min))            {                min = paths[0][i];                index = i;            }        }        if(min < 0)            break;        //cout<<"find point:"<<index<<",path="<<min<<endl;        shortest[len] = index;        len++;        //刷新各节点距离        for(i = 1; i < size; i++)        {            if(i == index)                continue;            if(paths[index][i] > 0)            {                temp = min+paths[index][i];                if(lessThen(temp, paths[0][i]))                {                    paths[0][i] = temp;                    keys[i] = index;                }            }            //cout<<"paths[0]["<<i<<"]="<<paths[0][i]<<endl;        }        cout<<endl;    }    delete shortest;    return keys;}bool lessThen(float x, float y){    if(x < 0)        return false;    if(y < 0)        return true;    return x < y;}

运行效果

带权有向图：

最短路径结果：