最短路打印路径
#include <iostream>
#include<string.h>
#include<stack>
#define M 100
#define N 100
using namespace std;
typedef struct node
{
??? int matrix[N][M];????? //鄰接矩陣
??? int n;???????????????? //頂點(diǎn)數(shù)
??? int e;???????????????? //邊數(shù)
}MGraph;
bool visited[1100];
void DijkstraPath(MGraph g,int *dist,int *path,int v0)?? //v0表示源頂點(diǎn)
{
??? int i,j,k;
?memset(visited,false,sizeof(visited));
??? for(i=0;i<g.n;i++)???? //初始化
??? {
??????? if(g.matrix[v0][i]>0&&i!=v0)
??????? {
??????????? dist[i]=g.matrix[v0][i];
??????????? path[i]=v0;???? //path記錄最短路徑上從v0到i的前一個(gè)頂點(diǎn)
??????? }
??????? else
??????? {
??????????? dist[i]=INT_MAX;??? //若i不與v0直接相鄰,則權(quán)值置為無窮大
??????????? path[i]=-1;
??????? }
??????? visited[i]=false;
??????? path[v0]=v0;
??????? dist[v0]=0;
??? }
??? visited[v0]=true;
??? for(i=1;i<g.n;i++)???? //循環(huán)擴(kuò)展n-1次
??? {
??????? int min=INT_MAX;
??????? int u;
??????? for(j=0;j<g.n;j++)??? //尋找未被擴(kuò)展的權(quán)值最小的頂點(diǎn)
??????? {
??????????? if(visited[j]==false&&dist[j]<min)
??????????? {
??????????????? min=dist[j];
??????????????? u=j;???????
??????????? }
??????? }
??????? visited[u]=true;
??????? for(k=0;k<g.n;k++)?? //更新dist數(shù)組的值和路徑的值
??????? {
??????????? if(visited[k]==false&&g.matrix[u][k]>0&&min+g.matrix[u][k]<dist[k])
??????????? {
??????????????? dist[k]=min+g.matrix[u][k];
??????????????? path[k]=u;
??????????? }
??????? }???????
??? }???
}
void showPath(int *path,int v,int v0)?? //打印最短路徑上的各個(gè)頂點(diǎn)
{
??? stack<int> s;
??? int u=v;
??? while(v!=v0)
??? {
??????? s.push(v);
??????? v=path[v];
??? }
??? s.push(v);
??? while(!s.empty())
??? {
??????? cout<<s.top()<<" ";
??????? s.pop();
??? }
}
int main()
{
??? int n,e;???? //表示輸入的頂點(diǎn)數(shù)和邊數(shù)
??? while(cin>>n>>e&&e!=0)
??? {
??????? int i,j;
??????? int s,t,w;????? //表示存在一條邊s->t,權(quán)值為w
??????? MGraph g;
??????? int v0;
??????? int *dist=(int *)malloc(sizeof(int)*n);
??????? int *path=(int *)malloc(sizeof(int)*n);
??????? for(i=0;i<N;i++)
??????????? for(j=0;j<M;j++)
??????????????? g.matrix[i][j]=0;
??????? g.n=n;
??????? g.e=e;
??????? for(i=0;i<e;i++)
??????? {
??????????? cin>>s>>t>>w;
??????????? g.matrix[s][t]=w;
??????? }
??????? cin>>v0;??????? //輸入源頂點(diǎn)
??????? DijkstraPath(g,dist,path,v0);
??????? for(i=0;i<n;i++)
??????? {
??????????? if(i!=v0)
??????????? {
??????????????? showPath(path,i,v0);
??????????????? cout<<dist[i]<<endl;
??????????? }
??????? }
??? }
??? return 0;
}
轉(zhuǎn)載于:https://www.cnblogs.com/lxm940130740/p/3278422.html
總結(jié)
- 上一篇: PL/SQL之高级篇
- 下一篇: 关于JAVA_HOME, CLASSPA