一：
#include "stdafx.h"
#include
#include
#include
using namespace std;
const int MAXINT = numeric_limits
::max();
template
void Dijkstra(int n, int v, Type dist[], int prev[], Type** c)
{
bool *s = new bool[n+1];
int i, j;
for(i = 1; i <=n; i++)
{
dist[i] = c[v][i];
if(c[v][i]!=MAXINT)
prev[i] = v;
else prev[i] =0;
s[i] = false;
}
s[v] = true;
dist[v] = 0;
prev[v] = 0;
for(i = 1; i< n; i++)
{
int u = v;
int temp = MAXINT;
for( j = 1; j<=n; j++)
if(!s[j]&&dist[j]
{
u = j;
temp = dist[j];
}
s[u] = true;
for(j=1; j<=n; j++)
{
if((!s[j])&&c[u][j]
{
if((dist[u]+c[u][j])
{
dist[j] = dist[u] + c[u][j];
prev[j] = u;
}
}
}
}
delete [] s;
}
void djpath(int m, int v, int prev[])
{
int i = m ,j =1;
while(i!=0)
{
if (j == 1)
{
cout<< i;
j = 0;
}
else
cout<< "-" << i;
i = prev[i];
}
cout<
}
int _tmain(int argc, _TCHAR* argv[])
{
cout< int prev[6], dist[6];
int i,j,n;
int** myc;
FILE *fp;
fp=fopen("data.txt", "r");
fscanf(fp,"%d", &n);
myc = new int* [n+1];
for(i =0; i<=n; i++)
myc[i] = new int[n+1];
for(i=1; i<=n; i++)
for(j =1; j<=n; j++)
{
fscanf(fp, "%d",&myc[i][j]);
if (myc[i][j]==-1)
myc[i][j] = MAXINT;
}
Dijkstra(5, 1, dist, prev, myc);
for(i = 1; i<=5; i++)
cout<
for(i=2; i<=5; i++)
djpath(i,1,prev);
for(i = 0; i<=5; i++)
delete[] myc[i];
delete [] myc;
return 0;
}
输入数据采用文本文件格式，下面是data.txt的内容:
5
0 10 -1 30 100
10 0 50 -1 -1
-1 50 0 20 10
30 -1 20 0 60
100 -1 10 60 0
本文来自CSDN博客，转载请标明出处：
二：
#include
using namespace std;
void main()
{
int infinity=100,j,i,n,k,t,**w,*s,*p,*d;
cout
cout<
d=new int[n];
s=new int[n];
p=new int[n];
w=new int*[n];
for(i=0;i
for(i=0;i
for(j=0;j
cin>>w[i][j];
for(s[0]=1,i=1;i
{
s[i]=0;d[i]=w[0][i];
if(d[i]
else p[i]=-1;
}
for(i=1;i
{
t=infinity;k=1;
for(j=1;j
if((!s[j])&&(d[j]
s[k]=1;//point k join the S
for (j=1;j
。且D[j]减小的新路径P只可能是由于路径
和边
组成。所以，当length(P)=D[k]+w
小于D[j]时，应该用P的长度来修改D[j]的值。
if((!s[j])&&(d[j]>d[k]+w[k][j]))
}
cout } 以上二个代码都是完整可以使用的，楼主喜欢哪个就用那个吧。自己还是可以仔细分析下代码。
