[2016-04-08] [POJ] [3159] [Candies]

• Time: 2016-04-08-18:05:09
• Title number: [2016-04-08][poj][3159][candies]
• The main topic: N Children, m information, each message A, B, C, said B children get sugar can not be more than a child C, ask N children than 1 children can more than the maximum number of
• Analysis: Direct is to ask for 1 to n the shortest path (if not the minimum, then the other side, there is always a situation can not meet the most short-circuit this side of the situation,), run directly once Dijkstra
• Problem encountered: Using ' vector ' to save the graph tle, changed to an array implementation of the adjacency table is a

  
 

   
  

  
#include <queue>
   
  

  
#include <algorithm>
   
  

  
#include <cstring>
   
  

  
#include <cstdio>
   
  

  
using namespace std;
   
  

  
const int maxn = 3 * 1E4 + 10;
   
  

  
const     int   maxe  =    15    *    1e4    +    10    
   
  

  
int mhead[maxn],nxt[maxe],mend[maxe],mcost[maxe],q;
   
  

  
int vis[maxn],d[maxn];
   
  

  
inline void addedge(int from,int to,int c){
   
  

  
 mend[q] = to;
   
  

  
 nxt[q] = mhead[from];
   
  

  
 mcost[q] = c;
   
  

  
 mhead[from] = q++;
   
  

  
}
   
  

  
inline void ini(){
   
  

  
 memset(mhead,-1,sizeof(mhead));
   
  

  
 q = 2;
   
  

  
}
   
  

  
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;
   
  

  
 }
   
  

  
};
   
  

  
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 e = mhead[u];~e;e = nxt[e]){
   
  

  
  int   v =   mend  [ e  ;   
   
  

  
  if   (! vis  [ v  ]    &&   D  [  v  ]    >   D  [ u      +   mcost  [ e     
   
  

  
 d[v] = d[u] + mcost[e];
   
  

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

  
 } 
   
  

  
 }
   
  

  
 }
   
  

  
}
   
  

  
int main(){
   
  

  
 int n,m,a,b,c;
   
  

  
 ini();
   
  

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

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

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

  
 addedge(a,b,c);
   
  

  
 }
   
  

  
 Dijkstra(1);
   
  

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

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-04-08] [POJ] [3159] [Candies]