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针对数学建模中的某一送货运输问题的C语言实现

附录:Floyd算法的C#实现
#include<stdio.h>
#define N 10
#define M 10000
void main()
{
	int r[N][N],i,j,k;
	int d[N][N]={0,50,M,40,25,M,30,M,50,M,
				50,0,30,M,35,50,M,60,M,M,
				M,30,0,15,M,30,50,25,M,60,
				40,M,15,0,45,30,55,20,40,65,
				25,15,M,45,0,60,10,30,M,55,
				M,50,30,30,60,0,25,55,35,M,
				30,M,50,M,10,25,0,30,45,60,
				M,60,25,20,30,55,30,0,10,M,
				20,M,M,40,M,15,25,45,0,20,
				35,20,10,45,20,M,60,M,30,0,};
	for(i=0;i<N;i++)
		for(j=0;j<N;j++)
			r[i][j]=j;
	for(k=0;k<N;k++)
		for(i=0;i<N;i++)
			for(j=0;j<N;j++)
				if(d[i][k]+d[k][j]<d[i][j]) 
{
					d[i][j]=d[i][k]+d[k][j];
					r[i][j]=k;
				}
	for(i=0;i<N;i++)
	{
		for(j=0;j<N;j++)
			printf("%3d ",d[i][j]);
		printf("\n");
	}
	printf("\n");
	for(i=0;i<N;i++)
	{
		for(j=0;j<N;j++)
			printf("%2d ",r[i][j]+1);
		printf("\n");
	}
}

"改良圈"算法的C#实现

#include<stdio.h>
#define N 10
#define M 10000
void main()
{
	int r[N][N],i,j,k,t,p,m,s=0;;
	int d[N][N]={0,50,M,40,25,M,30,M,50,M,
				50,0,30,M,35,50,M,60,M,M,
				M,30,0,15,M,30,50,25,M,60,
				40,M,15,0,45,30,55,20,40,65,
				25,15,M,45,0,60,10,30,M,55,
				M,50,30,30,60,0,25,55,35,M,
				30,M,50,M,10,25,0,30,45,60,
				M,60,25,20,30,55,30,0,10,M,
				20,M,M,40,M,15,25,45,0,20,
				35,20,10,45,20,M,60,M,30,0,},a[N][N],c[N]={0,4,6,1,3,8,2,5,7,9};
	for(i=0;i<N;i++)
		for(j=0;j<N;j++)
			r[i][j]=j;
	for(k=0;k<N;k++)
		for(i=0;i<N;i++)
			for(j=0;j<N;j++)
				if(d[i][k]+d[k][j]<d[i][j]) 
{
					d[i][j]=d[i][k]+d[k][j];
					r[i][j]=k;
				}
	for(i=0;i<N;i++)
		for(j=0;j<N;j++)
			a[i][j]=d[i][j];
	for(p=0;p<100;p++)
	{
		m=0;
		for(i=1;i<N;i++)
			for(j=i+2;j<N-1;j++)
				if(a[c[i]][c[j]]+a[c[i+1]][c[j+1]]<a[c[i]][c[i+1]]+a[c[j]][c[j+1]])
				{
					m=1;
					for(k=0;k<(j-i)/2;k++)
					{
						t=c[i+1+k];
						c[i+1+k]=c[j-k];
						c[j-k]=t;
					}
					for(i=0;i<N;i++)
						printf("%d ",c[i]+1);
					printf("\n");
				}
		
		if(m==0) break;
	}
}