数据结构 图 最短路径

（1）从某个源点到其余各顶点的最短路径

Dijkstra的代码如下：

头文件：


#define INFINITY 10000


#define MAX_VERTEX_NUM 20




typedef int InfoType;


typedef char VertexType;


typedef int VRType;






typedef enum ...{DG, DN, UDG, UDN} GraphKind;






typedef struct ArcCell...{


VRType adj;


InfoType *info;


}ArcCell, AdjMatrix[MAX_VERTEX_NUM][MAX_VERTEX_NUM];






typedef struct...{


VertexType vexs[MAX_VERTEX_NUM];


AdjMatrix arcs;


int vexnum, arcnum;


GraphKind kind;


}MGraph;






int weight[][6] = ...{




...{0, INFINITY, 10, INFINITY, 30, 100},




...{INFINITY, 0, 5, INFINITY, INFINITY, INFINITY},




...{INFINITY, INFINITY, 0, 50, INFINITY, INFINITY},




...{INFINITY, INFINITY, INFINITY, 0, INFINITY, 10},




...{INFINITY, INFINITY, INFINITY, 20, 0, 60},




...{INFINITY, INFINITY, INFINITY, INFINITY, INFINITY, 0}


};




//#define INPUT 1



源文件：


#include "graph_dj.h"


#include "stdio.h"


#include "stdlib.h"






void createDG(MGraph &g)...{}




void createDN(MGraph &g)...{}




void createUDG(MGraph &g)...{}






int locate(MGraph g, char name)...{




for(int i = 0; i < g.vexnum; i++)...{




if(name == g.vexs[i])...{


return i;


}


}


return -1;


}






void createUDN(MGraph &g)...{




#ifdef INPUT


printf("input the number of vertex and arcs: ");


scanf("%d %d", &g.vexnum, &g.arcnum);


fflush(stdin);


#else


printf("input the number of vertex: ");


scanf("%d", &g.vexnum);


fflush(stdin);


#endif




int i = 0, j = 0, k = 0;




printf("input the name of vertex: ");






for(i = 0; i < g.vexnum; i++)...{


scanf("%c", &g.vexs[i]);


fflush(stdin);


}






for(i = 0; i < g.vexnum; i++)...{




for(j = 0; j < g.vexnum; j++)...{


g.arcs[i][j].adj = INFINITY;


g.arcs[i][j].info = NULL;


}


}




#ifdef INPUT


char v1, v2;


int w;


printf("input the %d arcs v1 v2 and weight: ", g.arcnum);




for(k = 0; k < g.arcnum; k++)...{


scanf("%c %c %d", &v1, &v2, &w);


fflush(stdin);


i = locate(g, v1);


j = locate(g, v2);


g.arcs[i][j].adj = w;


}


#else


printf("now constructing the matrix of graph... ");




for(i = 0; i < g.vexnum; i++)...{




for(j = 0; j < g.vexnum; j++)...{


g.arcs[i][j].adj = weight[i][j];


}


}


#endif


}






void printGraph(MGraph g)...{




for(int i = 0; i < g.vexnum; i++)...{




for(int j = 0; j < g.vexnum; j++)...{


printf("%d ", g.arcs[i][j].adj);


}


printf(" ");


}


}






void createGragh(MGraph &g)...{




printf("please input the type of graph: ");


scanf("%d", &g.kind);






switch(g.kind)...{


case DG:


createDG(g);


break;


case DN:


createDN(g);


break;


case UDG:


createUDG(g);


break;


case UDN:


createUDN(g);


printGraph(g);


break;


}


}






void dijkstra(MGraph g)...{


int *dist = (int *)malloc(sizeof(int) * g.vexnum);


int *parent = (int *)malloc(sizeof(int) * g.vexnum);


int *final = (int *)malloc(sizeof(int) * g.vexnum);




int i, j, k, cost;


//init the arrays




for(i = 0; i < g.vexnum; i++)...{


dist[i] = g.arcs[0][i].adj;


parent[i] = 0;


final[i] = 0;


}


//join the start 0.


final[0] = 1;






for(i = 1; i < g.vexnum; i++)...{


//get the min of dist


cost = INFINITY;


k = -1;




for(j = 1; j < g.vexnum; j++)...{




if(dist[j] < cost && final[j] == 0)...{


cost = dist[j];


k = j;


}


}






if(k > 0)...{


//join the k node


final[k] = 1;




//update the dist




for(int t = 1; t < g.vexnum; t++)...{




if(final[t] == 0)...{




if(dist[t] > dist[k] + g.arcs[k][t].adj)...{


dist[t] = dist[k] + g.arcs[k][t].adj;


parent[t] = k;


}


}


}


}


}




//print the dist and parent array


printf("the shortest path from %c is as follows: ", g.vexs[0]);




for(i = 0; i < g.vexnum; i++)...{


printf("%c via %c's dist is %d ",


g.vexs[i], g.vexs[parent[i]], dist[i]);


}


}






void main()...{


MGraph g;


createGragh(g);


dijkstra(g);


}

程序的执行结果如下：


please input the type of graph:


3


input the number of vertex:


6


input the name of vertex:


a


b


c


d


e


f


now constructing the matrix of graph...


0 10000 10 10000 30 100


10000 0 5 10000 10000 10000


10000 10000 0 50 10000 10000


10000 10000 10000 0 10000 10


10000 10000 10000 20 0 60


10000 10000 10000 10000 10000 0


the shortest path from a is as follows:


a via a's dist is 0


b via a's dist is 10000


c via a's dist is 10


d via e's dist is 50


e via a's dist is 30


f via d's dist is 60


Press any key to continue

说明：Dijkstra算法的复杂度是O(n * n)。

（2）每一对顶点之间的最短路径

方法一：对每个顶点依次运行Dijkstra，复杂度O(n*n*n)。

方法二：floyd，复杂度O(n*n*n)。

头文件修改：




int weight[][3] = ...{




...{0, 4, 11},




...{6, 0, 2},




...{3, INFINITY, 0}


};

源文件添加：




void floyd(MGraph g)...{


int dist[MAX_VERTEX_NUM][MAX_VERTEX_NUM];


int parent[MAX_VERTEX_NUM][MAX_VERTEX_NUM];




//init the arrays


int i, j;




for(i = 0; i < g.vexnum; i++)...{




for(j = 0; j < g.vexnum; j++)...{


dist[i][j] = g.arcs[i][j].adj;


parent[i][j] = i;


}


}




//compute the shortest path




for(int k = 0; k < g.vexnum; k++)...{




for(i = 0; i < g.vexnum; i++)...{




for(j = 0; j < g.vexnum; j++)...{




if(i != j && i != k && j != k)...{




if(dist[i][j] > dist[i][k] + dist[k][j])...{


dist[i][j] = dist[i][k] + dist[k][j];


parent[i][j] = k;


}


}


}


}


}




//print the shortest path and parent




for(i = 0; i < g.vexnum; i++)...{




for(j = 0; j < g.vexnum; j++)...{


printf("%d(%c) ", dist[i][j], g.vexs[parent[i][j]]);


}


printf(" ");


}


}






void main()...{


MGraph g;


createGragh(g);


dijkstra(g);


floyd(g);


}

程序的运行结果如下：


please input the type of graph:


3


input the number of vertex:


3


input the name of vertex:


a


b


c


now constructing the matrix of graph...


0 4 11


6 0 2


3 10000 0


the shortest path from a is as follows:


a via a's dist is 0


b via a's dist is 4


c via b's dist is 6


0(a) 4(a) 6(b)


5(c) 0(b) 2(b)


3(c) 7(a) 0(c)


Press any key to continue