弗洛伊德算法计算所有顶点对之间最短通路的长度

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//弗洛伊德算法计算所有顶点对之间最短通路的长度// procedure Floyd(G:带权简单图)//  {G有顶点v1.v2....vn和权}// for i= 1 to n///  for j= 1 to n//     d(vi,vj)=w（i，j）//// for i= 1 to n///  for j= 1 to n//     for k= 1 to n//       if d(vj,vi)+d(vi,vk)<d(vj,vk)//        then d(vj,vk) = d(vj,vi)+d(vi,vk)// d(vi,vj)是在vi到vj之间的最短通路//这个算法是计算所有顶点对之间最短通路的长度//具体实现，应该要用到保存所有顶点对之间最短通路的长度的存贮单位(这里使用了二维数组存贮)/*粘贴输入下列测试数据  6 90 1 40 2 21 2 11 3 52 3 82 4 103 4 23 5 64 5 3'''''''''''''''''''''''''''''''''''''''''''''''''0 4 2 32767 32767 327674 0 1 5 32767 327672 1 0 8 10 3276732767 5 8 0 2 632767 32767 10 2 0 332767 32767 32767 6 3 00 ,0:00 ,1:30 ,2:20 ,3:80 ,4:100 ,5:131 ,1:01 ,2:11 ,3:51 ,4:71 ,5:102 ,2:02 ,3:62 ,4:82 ,5:113 ,3:03 ,4:23 ,5:54 ,4:04 ,5:35 ,5:0Press any key to continue*/#include <iostream>using namespace std;//012345分别表示v0 v1......  int v;//////点数目  int edge;///边数目int** draw(int ** &l);//绘图void shuchu(int **l);//输出int main(int argc, char const *argv[]){   int **l;  int**a=draw(l);//------------------------------------------------------------------------////------------------------------------------------------------------------//  int i,j,k;  for ( i = 0; i < v; ++i)  for ( j = 0; j < v; ++j)  for (k = 0; k < v; ++k)             if (l[j][i] + l[i][k] < l[j][k]) l[j][k] = l[j][i] + l[i][k];   //------------------------------------------------------------------------//  shuchu(l);  return 0;}int** draw(int ** &l)//绘图{  int i,j;  cin>>v>>edge;///输入点数目和边数目  int **a=new int*[v];  for ( i = 0; i < v; ++i)    a[i]=new int[v];  l=new int*[v];   for ( i = 0; i < v; ++i)     l[i]=new int[v];  for ( i = 0; i < v; ++i)    for ( j = 0; j < v; ++j)      {if(i==j){a[i][j]=0;l[i][j]=0;continue;}a[i][j]=32767;l[i][j]=32767;}  int spot1,spot2,len;  for ( i = 0; i < edge; ++i)  {    cin>>spot1>>spot2>>len;    a[spot1][spot2]=len;    a[spot2][spot1]=len;     l[spot1][spot2]=len;    l[spot2][spot1]=len;  }///-------------------------------------------------------------///    for ( i = 0; i < v; ++i){    for ( j = 0; j < v; ++j)      cout<<a[i][j]<<" ";    cout<<"\n";    }    cout<<"\n";  return a;}void shuchu(int **l){int i,j;   for ( i = 0; i < v; ++i)      for ( j = i; j < v; ++j)        cout<<i<<" ,"<<j<<":"<<l[i][j]<<"\n";}
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