[2016-04-02] [POJ] [2387] [Til the Cows Come Home]

• Time: 2016-04-02-10:34:36 Saturday
• Title Number: [2016-04-02][poj][2387][til the Cows Come Home]
• Topic: given n nodes and T-path, the shortest-circuit length of N to 1 is obtained.
• Analysis: Run the shortest possible time
• Problems encountered:
  • is said to be multiple sides, if the adjacency matrix is used to update the minimum value,
  • This problem is first input t, and then enter N, input the time to read wrong, infinite wa ...

  
 

   
  

  
#include <queue>
   
  

  
#include <cstring>
   
  

  
#include <cstdio>
   
  

  
using namespace std;
   
  

  
const int maxn = 1000 + 10;
   
  

  
const int maxe = 2000 + 10;
   
  

  
struct mNode{
   
  

  
 int u,c;
   
  

  
 Mnode   ( int   _u  =    0   int   _c  =    0  ):  u   ( _u   c   ( _c    
   
  

  
 bool operator < (const mNode & a)const{
   
  

  
 return c > a.c;
   
  

  
 }
   
  

  
};
   
  

  
struct Edge{
   
  

  
 int v,c;
   
  

  
 Edge(int _v = 0,int _c = 0):v(_v),c(_c){}
   
  

  
};
   
  

  
vector<Edge> e[maxe];
   
  

  
bool vis[maxn];
   
  

  
int d[maxn];
   
  

  
void Dijkstra(int s){
   
  

  
 memset(vis,0,sizeof(vis));
   
  

  
 memset(d,0x3f,sizeof(d));
   
  

  
 priority_queue<mNode> q;
   
  

  
 d[s] = 0;
   
  

  
 q.push(mNode(s,0));
   
  

  
 mNode tmp;
   
  

  
 while(!q.empty()){
   
  

  
 tmp = q.top();
   
  

  
 q.pop();
   
  

  
 int u = tmp.u;
   
  

  
 if(vis[u]) continue;
   
  

  
 vis[u] = 1;
   
  

  
 for(int i = 0;i < e[u].size();++i){
   
  

  
 Edge & me = e[u][i];
   
  

  
  if   (! vis  [ me   v      &&   D  [ me   v      >   D  Span class= "pun" >[ u  ]    +   me   c     
   
  

  
 D  [me   v      =   D  [ u  ]    +   me   c    
   
  

  
 q.push(mNode(me.v,d[me.v]));
   
  

  
 } 
   
  

  
 }
   
  

  
 }
   
  

  
}
   
  

  
inline void addedge(int u,int v,int c){
   
  

  
 e[u].push_back(Edge(v,c));
   
  

  
}
   
  

  
int main(){
   
  

  

   
  

  
 int n,t,a,b,c;
   
  

  
 scanf("%d%d",&t,&n);
   
  

  
 for(int i = 0;i < t; ++i){
   
  

  
 scanf("%d%d%d",&a,&b,&c);
   
  

  
 addedge(a,b,c);
   
  

  
 addedge(b,a,c);
   
  

  
 }
   
  

  
 Dijkstra(n);
   
  

  
 printf("%d\n",d[1]);
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-02] [POJ] [2387] [Til the Cows Come Home]