对dijkstra算法理解有帮助的文章

 

http://www.cnblogs.com/SadGeminids/archive/2011/12/29/2306719.html

 

http://wenku.baidu.com/view/1e9c73ef4afe04a1b071ded1.html

第二个当题做

#include <stdio.h>

#define true  1

#define false 0

#define I  9999                                /* 无穷大        */

#define N  20                                  /* 城市顶点的数目 */

int cost[N][N] = {

{0,3,I,I,I,1,I,I,I,I,I,I,I,I,I,I,I,I,I,I},

{3,0,5,I,I,I,6,I,I,I,I,I,I,I,I,I,I,I,I,I},

{I,5,0,4,I,I,I,1,I,I,I,I,I,I,I,I,I,I,I,I},

{I,I,4,0,2,I,I,I,6,I,I,I,I,I,I,I,I,I,I,I},

{I,I,I,2,0,I,I,I,I,7,I,I,I,I,I,I,I,I,I,I},

{1,I,I,I,I,0,1,I,I,I,2,I,I,I,I,I,I,I,I,I},

{I,6,I,I,I,1,0,6,I,I,I,7,I,I,I,I,I,I,I,I},

{I,I,1,I,I,I,6,0,2,I,I,I,3,I,I,I,I,I,I,I},

{I,I,I,6,I,I,I,2,0,8,I,I,I,4,I,I,I,I,I,I},

{I,I,I,I,7,I,I,I,8,0,I,I,I,I,5,I,I,I,I,I},

{I,I,I,I,I,2,I,I,I,I,0,4,I,I,I,3,I,I,I,I},

{I,I,I,I,I,I,7,I,I,I,4,0,3,I,I,I,4,I,I,I},

{I,I,I,I,I,I,I,3,I,I,I,3,0,3,I,I,I,5,I,I},

{I,I,I,I,I,I,I,I,4,I,I,I,3,0,7,I,I,I,2,I},

{I,I,I,I,I,I,I,I,I,5,I,I,I,7,0,I,I,I,I,3},

{I,I,I,I,I,I,I,I,I,I,3,I,I,I,I,0,5,I,I,I},

{I,I,I,I,I,I,I,I,I,I,I,4,I,I,I,5,0,8,I,I},

{I,I,I,I,I,I,I,I,I,I,I,I,5,I,I,I,8,0,6,I},

{I,I,I,I,I,I,I,I,I,I,I,I,I,2,I,I,I,6,0,4},

{I,I,I,I,I,I,I,I,I,I,I,I,I,I,3,I,I,I,4,0}

};

int dist[N];                                         /* 存储当前最短路径长度 */

int v0 ='A' - 65;                                   /* 初始点是 A          */

void main()

{

intfinal[N], i, v, w, min;

/* 初始化最短路径长度数据，所有数据都不是最终数据 */

for(v = 0; v < N; v++) {

final[v] =false;

dist[v] = cost[v0][v];

}

/* 首先选v0到v0的距离一定最短，最终数据 */

final[v0] =true;

/* 寻找另外 N-1 个结点 */

for(i = 0; i < N-1; i++) {

min = I;                                     /* 初始最短长度无穷大  */

/* 寻找最短的边 */

for(w = 0; w < N; w++) {

if(!final[w] && dist[w] < min) {

min = dist[w];

v = w;

}

}

final[v] =true;                             /* 加入新边          */

for(w = 0; w < N; w++) {                      /* 更新 dist[] 数据  */

if(!final[w] && dist[v] + cost[v][w] < dist[w]) {//遍历所有可能了

dist[w] = dist[v] + cost[v][w];

}

}

}

for(i = 0; i < N; i++) {                          /* 显示到监视器      */

printf("%c->%c: %2d\t", v0 + 65, i + 65, dist[i]);

}

}