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数据结构----单源最短路径Dijkstra

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原理：参考趣学数据结构

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代码：

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stack.h 栈代码

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#pragma once
#include<stdio.h>
#define maxSize 100
typedef struct stack {int * base;int * top;
}stack;
bool init(stack & Stack) {//栈的初始化Stack.base = new int[maxSize];if (!Stack.base) {return false;}Stack.top = Stack.base;return true;
}
bool push(stack & Stack,int e) {//入栈if (Stack.top - Stack.base == maxSize) {//满栈，不能再插入return false;}*(Stack.top) = e;Stack.top++;return true;
}
bool pop(stack & Stack, int &e) {//出栈if (Stack.base == Stack.top) {//栈空return false;}Stack.top--;e = *(Stack.top);return true;
}
int getTop(stack &Stack) {if (Stack.base == Stack.top) {//栈空return -1;}return *(Stack.top - 1);
}
void printStack(stack Stack) {//遍历栈中的元素while (Stack.base != Stack.top) {printf("%d ", *(Stack.top - 1));Stack.top--;}
}
bool empty(stack Stack) {//栈空的判断if (Stack.base == Stack.top) {return true;}return false;
}
void testStack() {//测试栈是否有问题stack Stack;init(Stack);int value;while (1) {scanf_s("%d", &value);if (value == -1) {break;}push(Stack, value);}printStack(Stack);
}

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dijkstra.cpp

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#include<stdio.h>
#include<stdlib.h>
#include"stack.h"
#define N 100
#define elemType int
//const int MAX_INT = (1 << 31) - 1;
//const int MAX_INT = 0X7fffffff;
#define  INF    (((unsigned int)(-1)) >> 1)
typedef struct GraphMatrix {elemType vNode[N][N];int vNum, eNum;
}GraphMatrix;
void findPath(GraphMatrix G, int dist[], int p[], int u, stack &Stack);//声明
void initGMaxtix(GraphMatrix &G) {//初始化邻接矩阵printf("输入顶点数和边数\n");scanf_s("%d%d", &G.vNum, &G.eNum);for (int i = 0; i < G.vNum; i++) {//初始化邻接矩阵for (int j = 0; j < G.vNum; j++) {G.vNode[i][j] = G.vNode[j][i]=INF;}}printf("输入顶点v1到顶点v2和其边的权重\n");for (int i = 0; i < G.eNum; i++ ) {int v1, v2,weights;scanf_s("%d%d%d", &v1, &v2,&weights);G.vNode[v1][v2] = G.vNode[v2][v1] = weights;}
}
void print4(GraphMatrix G) {printf("邻接矩阵如下:\n");for (int i = 0; i < G.vNum; i++) {for (int j = 0; j < G.vNum; j++) {printf("%d ", G.vNode[i][j]);}printf("\n");}
}
void print41(int result[], int length) {for (int i = 0; i < length; i++) {printf("%d ", result[i]);}printf("\n");
}
void Dijkstra(GraphMatrix G, int u) {//单源最短路径，分为两个集合求解bool flag[N];//标识是否加入第一个顶点集int dist[N], p[N];//分别为单源点到其他点的距离和这段距离所经过的那些顶点的记录数组pp[u] = -1;dist[u] = 0;for (int i = 0; i < G.vNum; i++) {flag[i] = false;//每一个顶点都默认没有访问dist[i] = G.vNode[u][i];if (G.vNode[u][i]==INF) {//对距离数组和记录数组初始化p[i] = -1;}else {p[i] = u;}}flag[u] = true;for (int i = 0; i < G.vNum - 1; i++) {//G.vNum-1次连接//找没有被访问的最小元素int min=INF;int t=u;for (int j = 0; j < G.vNum; j++) {if (!flag[j] && dist[j] < min) {t = j;min = dist[j];}}if (t == u) {break;//全部顶点加入到第一个顶点集合中}flag[t] = true;//加入到第一个顶点集合中//更新距离数组for (int j = 0; j < G.vNum; j++) {if (!flag[j] && dist[t] + G.vNode[t][j] < dist[j]) {dist[j] = dist[t] + G.vNode[t][j];p[j] = t;}}}//print41(dist, G.vNum);stack Stack;init(Stack);printf("输出单源最短路径的最优方案\n");findPath(G, dist, p, u, Stack);
}
void findPath(GraphMatrix G,int dist[], int p[],int u,stack &Stack) {for (int i = 0; i < G.vNum; i++) {if (p[i]==-1) {//起点到起点printf("%d---%d不可达!\n",u,i);continue;}push(Stack, i);int x = p[i];while (x != -1) {//入栈查找路径push(Stack, x);x = p[x];}int e;while (!empty(Stack)) {//出栈遍历路径printf("%d", getTop(Stack));pop(Stack, e);if (Stack.top - Stack.base >= 1) {printf("---");}}printf(" 这段路径的距离为:%d\n", dist[i]);}
}
int main() {GraphMatrix G;initGMaxtix(G);print4(G);printf("\n");Dijkstra(G,0);system("pause");return 0;
}

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测试截图：

时间复杂度为O(n)，空间复杂度为O(n)!

如果存在什么问题，欢迎批评指正！谢谢！