最短路径（弗洛伊德算法）

Posted 2020-10-06 不二小杰

假设条件同上。。

整个算法最核心的，个人觉得就是一个公式：

weight[a][b] = min{weight[a][b], weight[a][c]+weight[c][b]}

即，从一点到另外一点的最短距离，是在直线和经过一个中间点的‘绕路’距离之间求最短。。然后利用上一次的结果迭代。。

/*
*  author: buer
*  github: buer0.github.com
*/
#include <stdio.h>
#include <stdlib.h>

#define MAXSIZE 10

typedef struct Graph
{
    int table[MAXSIZE][MAXSIZE];
    int num;
}Graph;

void createTable(Graph *graph);
void printTable(Graph *graph); 
void shortest(Graph *graph);
 
int main(int argc, char *argv[])
{
    Graph graph;
    
    
    createTable(&graph);
    
    printTable(&graph);
    
    shortest(&graph); 
    return 0;
}

void shortest(Graph *graph) 
{
    int num = graph->num;
    int pre[num][num]; 
    int weight[num][num];
    int i, j, k;
    
    for(i=0; i<num; i++)
    {
        for(j=0; j<num; j++)
        {
            pre[i][j] = j;
            weight[i][j] =  (graph->table)[i][j];
        }
    }
    
    
    for(i=0; i<num; i++)
    {
        for(j=0; j<num; j++)
        {
            for(k=0; k<num; k++)
            {
                if( weight[i][k] > weight[i][j] + weight[j][k] )
                {
                    weight[i][k] = weight[i][j] + weight[j][k];
                    pre[i][k] = j;
                }
            }
            
        }
        
    }
    
    printf("result:\n");
    for(i=0; i<num; i++)
    {
        for(j=0; j<num; j++)
        {
            printf("%d ", weight[i][j]);
        }
        printf("\n");
    }
    
    
}

void createTable(Graph *graph)
{
    int i, j, temp;
    printf("输入节点数:");
    scanf("%d", &(graph->num));
    
    for(i=0; i<graph->num; i++)
    {
        printf("第 %d 行:", i+1);
        for(j=0; j<graph->num; j++)
        {
            scanf("%d", &temp);
            if(temp == ‘ ‘)
            {
                j --;
            }else {
                (graph->table)[i][j] = temp;
            }
        }
        getchar();
    }
}

void printTable(Graph *graph)
{
    int i,j;
    printf("\n");
    for(i=0; i<graph->num; i++)
    {
        for(j=0; j<graph->num; j++)
        {
            printf("%d ", (graph->table)[i][j]);
        }
        printf("\n");
    }
 }