tu.cpp

按动态规划余利解决计算图的任意两点见的距离的Floyd算法

#include "iostream.h"
#include "stdio.h"
#define n 10

void main()
{
	int i,j,k;
	int c[n][n]={{0,99,8,7,6,5,4,3,2,1},{99,0,99,8,7,6,5,4,3,2},
	{8,99,0,99,8,7,6,5,4,3},{7,8,99,0,99,8,7,6,5,4 },
	{6,7,8,99,0,99,8,7,6,5},{5,6,7,8,99,0,99,8,7,6},
	{4,5,6,7,8,99,0,99,8,7},{3,4,5,6,7,8,99,0,99,8},
	{2,3,4,5,6,7,8,99,0,99},{1,2,3,4,5,6,7,8,99,0}};
	int a[n][n];//,b[n]={0,99,8,7,6,5,4,3,2,1};
	//for(i=0;  i<n;  i++)
	//	for (j=0,k=0;  j<n;  j++)
		//{
		//	c[i][j]=b[k++];	
		//	if(i=j) c[i][j]=0;
		//	if(k=10) k=0;
	//	};//输入
		for(i=0; i<n; i++)
			for(j=0; j<n; j++)
				a[i][j]=c[i][j];

		for (i=0; i<n; i++)
			for(j=0; j<n; j++)
			{
				int *p;
				p=&c[0][0];
				cout<<*p++<<"  ";
				//printf("%d  ",c[i][j]);
				//if(j==9) printf("\n");
			}//输出c[][]
			printf("\n");
		for (k=0; k<n; k++)
			for(i=0; i<n; i++)
				for(j=0;  j<n; j++)
				{
					if( a[i][j]>a[i][k]+a[k][j] )
						a[i][j]=a[i][k]+a[k][j];
				}//查找
	//	for (i=0; i<n; i++)
		//	for(j=0; j<n; j++)
		//	{
		//		printf("%d  ",a[i][j]);
		//		if(j==9) printf("\n");
		//	}//输出a[][]



}