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数据结构—多源最短路径

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原理：参考趣学数据结构

代码：

栈代码：

#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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多源最短路径代码：

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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[N][N], int p[N][N],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]= 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]  = weights;}
}
void print10(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 print101(int result[N][N], int length) {for (int i = 0; i < length; i++) {for (int j = 0; j < length; j++) {printf("%d ", result[i][j]);}printf("\n");}printf("\n");
}
void Floyd(GraphMatrix G) {//多源最短路径求解int dist[N][N], p[N][N];//分别为多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pfor (int i = 0; i < G.vNum; i++) {for (int j = 0; j < G.vNum; j++) {dist[i][j] = G.vNode[i][j];if (G.vNode[i][j] == INF || i==j ) {//对距离数组和记录数组初始化p[i][j] = -1;}else {p[i][j] = i;}}}for(int k=0;k<G.vNum;k++){for (int i = 0; i < G.vNum; i++) {//更新距离数组for (int j = 0; j < G.vNum; j++) {//注意INF+INF越界问题  如果没有发现多调试if (dist[i][k]<INF && dist[k][j]<INF && dist[i][k] + dist[k][j] < dist[i][j]) {dist[i][j] = dist[i][k] + dist[k][j];p[i][j] = p[k][j];}}}}//print101(dist, G.vNum);stack Stack;init(Stack);printf("输出多源最短路径的最优方案\n");findPath(G, dist, p,Stack);
}
void findPath(GraphMatrix G, int dist[N][N], int p[N][N], stack &Stack) {for (int i = 0; i < G.vNum; i++) {for (int j = 0; j < G.vNum; j++) {if (p[i][j] == -1) {//起点到起点printf("%d---%d不可达!\n", i, j);continue;}push(Stack, j);int x = p[i][j];while (x != -1) {//入栈查找路径push(Stack, x);x = p[i][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][j]);}}
}
int main() {GraphMatrix G;initGMaxtix(G);print10(G);printf("\n");Floyd(G);system("pause");return 0;
}

测试截图：

时间复杂度为O(n x n x n)，空间复杂度为O(n x n)

如果存在什么问题，欢迎批评指正！谢谢！