最短路算法(写烂它，倒背如流）

Floyd算法一定要记住，探测点循环要放到最外层！！！！！！！！！！！！！！！

这是由其算法本身所决定的，其每一步求出任意一对顶点之间仅通过中间节点1,2,...,k的最短距离，当1,2,...,k扩展到所有顶点时，算法解出任意一对顶点间的最短距离，故顺序自然是：for(k=1;k<n;++k)    //枚举任意一对顶点由其状态转移方程来看，这个算法的顺序也很清晰，应该是先计算较小的k时任意ij之间的最短距离：dij(k) = wij  如果k=0min(dij(k-1),dik(k-1)+dkj(k-1))   如果k>=1其中i,j表示点对，k表示第1,2,...,k时的最短路径

const maxV = 100;  const maxLen = 999999;    int dist[maxV];  int path[maxV];  int length[maxV][maxV];  int i,j;      void Dijkstra(int v,int n)  {      bool s[maxV];      for(i=1;i<=n;++i)      {          s[i]=0;          dist[i]=length[v][i];          if(dist[i]==maxLen)              path[i]=0;          else              path[i]=v;      }      s[v]=1;      dist[v]=0;      for(i=2;i<=n;++i)      {          int temp = maxLen;          int u = v;          for(j=1;j<=n;++j)          {              if(!s[j]&&dist[j]<temp)              {                  u = j ;                  temp = dist[j];              }          }          s[u]=1;          for(j=1;j<=n;j++)          {              if(!s[j]&&dist[j]>length[u][j]+dist[u])              {                  path[j]=u;                  dist[j]=length[u][j]+dist[u];              }          }      }      }  void floyd()  {  for(k=1;k<=n;k++)     for(i=1;i<=n;i++)          for(j=1;j<=n;j++)                  if(length[i][k]+length[k][j]<length[i][j])                      length[i][j]=length[i][k]+length[k][j];  }

Bellman-Ford 算法

#include "stdio.h"const MaxLen=999999;const MaxV = 1000;const MaxE=1000;int n,e;int dist[MaxV];int i,j;struct edge{int u;int v;int weight;}edge[MaxE];bool bellman_ford(){for(i=1;i<n;i++){for(j=1;j<=e;j++){if(dist[edge[j].u]+edge[j].weight<dist[edge[j].v])dist[edge[j].v]=dist[edge[j].u]+edge[j].weight;}}for(j=1;j<=e;j++){if(dist[edge[j].u]+edge[j].weight<dist[edge[j].v])return false;}return true;}int main(int argc, char* argv[]){while(scanf("%d%d",&n,&e)!=EOF){for(i= 1;i<=e;i++){scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].weight);if(edge[i].u==1)dist[edge[i].v]=edge[i].weight;elsedist[edge[i].v]=MaxLen;}bellman_ford();for(i= 2;i<=n;i++){printf("到%d点 %d\n",i,dist[i]);}}return 0;}