luogu 4366 [Code+#4]最短路 Dijkstra + 位运算 + 思维

Posted 2022-08-12 guangheli

Code: 

#include <cstdio> 
#include <algorithm>
#include <cstring>  
#include <queue> 
#define N 100004    
#define M 4000000 
#define inf 10000000000000 
#define ll long long 
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std;  
struct Node 

    int u; 
    ll dis; 
    Node(int u=0,ll dis=0):u(u),dis(dis)   
    bool operator<(Node b)const
    
        return b.dis<dis;   
     
;  
priority_queue<Node>q;   
int n,m,C,edges,s,t; 
ll d[N];    
int hd[N],nex[M],to[M],val[M],done[N];   
inline void addedge(int u,int v,int c) 
 
    nex[++edges]=hd[u],hd[u]=edges,to[edges]=v,val[edges]=c;  
 
inline void Dijkstra() 
 
    int i,u,v; 
    for(i=0;i<=n;++i) d[i]=inf;
    d[s]=0, q.push(Node(s,0));  
    while(!q.empty()) 
    
        Node e=q.top(); u=e.u,q.pop();
        if(done[u]) continue;  
        done[u]=1;
        for(i=hd[u];i;i=nex[i]) 
            if(d[to[i]]>d[u]+val[i]) 
                d[to[i]]=d[u]+val[i],q.push(Node(to[i],d[to[i]]));    
          

int main() 

    int i,j; 
    // setIO("input"); 
    scanf("%d%d%d",&n,&m,&C);
    for(i=1;i<=m;++i) 
     
        int u,v,c; 
        scanf("%d%d%d",&u,&v,&c), addedge(u,v,c);       
     
    for(i=0;i<=n;++i) 
    
        int l; 
        for(l=0;(1<<l)<=n;++l) if((i^(1<<l))<=n)addedge(i,i^(1<<l),(1<<l)*C);          
      
    scanf("%d%d",&s,&t),Dijkstra(),printf("%lld\n",d[t]); 
    return 0;