标签:void 邻接矩阵 i++ ++ main 代码 sele oid include
使用邻接矩阵存储加权图,无穷大使用常数MAXLEN代表,然后使用Dijkstra方法求取最短路径
1 #include <stdio.h> 2 3 #define MAXLEN 1000 4 int cost[7][7]; 5 int dist[7]; 6 7 void creategraph(int* node,int num) 8 { 9 int from; 10 int to; 11 int i; 12 for(i = 0;i < num; i++) 13 { 14 from = node[i*3]; 15 to = node[i*3+1]; 16 cost[from][to] = node[i*3+2]; 17 } 18 } 19 20 void shortestpath(int begin,int num) 21 { 22 int selected[7]; 23 int min; 24 int s; 25 int i,j; 26 27 for(i = 2;i <= num;i++) 28 { 29 selected[i] = 0; 30 dist[i] = cost[begin][i]; 31 } 32 selected[begin] = 1; 33 dist[begin] = 0; 34 printf("顶点1 2 3 4 5 6\n"); 35 for(j = 1;j <= num;j++) 36 printf(" %4d ",dist[j]); 37 printf("\n"); 38 for(i = 1;i <= num - 1;i++) 39 { 40 min = MAXLEN; 41 for(j = 1;j <= num;j++) 42 if(min > dist[j] && selected[j] == 0) 43 { 44 s = j; 45 min = dist[j]; 46 } 47 selected[s] = 1; 48 for(j = 1;j <= num;j++) 49 { 50 if(selected[j] == 0 && 51 dist[s] + cost[s][j] < dist[j]) 52 dist[j] = dist[s] + cost[s][j]; 53 printf(" %4d ",dist[j]); 54 } 55 printf("\n"); 56 } 57 } 58 59 int main() 60 { 61 int node[7][3] = { 62 { 1, 2, 35}, 63 { 2, 3, 45}, 64 { 2, 4, 30}, 65 { 3, 5, 25}, 66 { 4, 5, 45}, 67 { 4, 6, 130}, 68 { 5, 6, 100}, 69 }; 70 int i,j; 71 72 for(i = 1;i <= 6;i++) 73 for(j = 1;j <= 6;j++) 74 cost[i][j] = MAXLEN; 75 creategraph(node,7); 76 printf("加权图的邻接矩阵内容:\n"); 77 for(i = 1;i <= 6;i++) 78 { 79 for(j = 1;j <= 6;j++) 80 printf(" %4d ",cost[i][j]); 81 printf("\n"); 82 } 83 printf("\n从顶点1到各顶点最近距离计算过程:\n"); 84 shortestpath(1,6); 85 }
图的最短路径(Dijkstra)(Floyd),拓扑排序,生成树代码
标签:void 邻接矩阵 i++ ++ main 代码 sele oid include
原文地址:https://www.cnblogs.com/hulianxingkong/p/14224228.html