Luogu P4366 [Code+#4]最短路 题解

#include<bits/stdc++.h>
using namespace std;
inline int read(){
int res=0,f=1;char ch=getchar();
while(!isdigit(ch)) f=ch=='-'?-1:f,ch=getchar();
while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch&15),ch=getchar();
return res*f;
}
inline void write(int x){
if(x<0) putchar('-'),x=-x;
if(x<10) putchar(x+'0');
else write(x/10),putchar(x%10+'0');
}
int n,m,c,fir[100010],nxt[500010+20*100010],son[500010+20*100010],w[500010+20*100010],tot,A,B,vis[1000010],dis[1000010];
inline void add(int x,int y,int z){++tot;nxt[tot]=fir[x];fir[x]=tot;son[tot]=y;w[tot]=z;}
inline void build(){
for(int i=0;i<=n;i++)
for(int j=1;j<=n;j<<=1)
if((i^j)<=n) add(i,i^j,j*c);
}
priority_queue<pair<int,int> > q;
inline void Dij(){
while(!q.empty()) q.pop();
q.push(make_pair(0,A));
memset(dis,63,sizeof(dis));dis[A]=0;
memset(vis,0,sizeof(vis));
while(!q.empty()){
pair<int,int> u=q.top();q.pop();
while(vis[u.second]&&!q.empty()) u=q.top(),q.pop();
if(vis[u.second]) return ;
vis[u.second]=1;
for(int to,i=fir[u.second];i;i=nxt[i]){
to=son[i];
if(dis[to]>dis[u.second]+w[i]){
dis[to]=dis[u.second]+w[i];
q.push(make_pair(-dis[to],to));
}
}
}
}
int main(){
n=read(),m=read(),c=read();
build();
for(int x,y,z,i=1;i<=m;i++) x=read(),y=read(),z=read(),add(x,y,z);
A=read(),B=read();
Dij();
return write(dis[B]),0;
}