bzoj3040: 最短路(road)(dijkstra)
阿新 • • 發佈:2019-02-11
Solution
配對堆優化
Code
#include<bits/stdc++.h>
#include<ext/pb_ds/priority_queue.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<ll,int> pa;
typedef __gnu_pbds::priority_queue<pa,greater<pa>,pairing_heap_tag> heap;
#define mp make_pair
const int N=1000001;
struct node{
int to,ne,w;
}e[10000001];
heap q;
heap::point_iterator id[N];
ll dis[N];
int n,m,T,rxa,rxc,rya,ryc,rp,a,b,x,y,z,i,tot,h[N],u;
inline char gc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1== p2)?EOF:*p1++;
}
inline int read(){
int x=0,fl=1;char ch=gc();
for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
return x*fl;
}
void add(int x,int y,int z){
e[++tot]=(node){y,h[x],z};
h[x]=tot;
}
int main (){
n=read();m=read();T=read();rxa=read();rxc=read();rya=read();ryc=read();rp=read();
for (i=0;i<T;i++){
x=(1ll*x*rxa+rxc)%rp;
y=(1ll*y*rya+ryc)%rp;
b=y%n+1;
a=min(x%n+1,b);
add(a,b,100*(1e6-a));
}
for (i=0;i<m-T;i++){
x=read();y=read();z=read();
add(x,y,z);
}
memset(dis,63,(n+1)<<3);
dis[1]=0;
q.push(mp(0,1));
while (!q.empty()){
u=q.top().second;if (u==n) break;q.pop();
for (int i=h[u],v;i;i=e[i].ne){
v=e[i].to;
if (dis[u]+e[i].w<dis[v]){
dis[v]=dis[u]+e[i].w;
if (id[v]==0) id[v]=q.push(mp(dis[v],v));
else q.modify(id[v],mp(dis[v],v));
}
}
}
printf("%lld",dis[n]);
}
二叉堆(洛谷P4779):
#include<bits/stdc++.h>
using namespace std;
const int N=100002;
struct node{
int to,ne,w;
}e[N<<1];
struct kk{
int u,d;
bool operator <(const kk &x)const{return d>x.d;}
};
int n,m,s,i,dis[N],h[N],tot,x,y,z;
priority_queue<kk>q;
inline char gc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
int x=0,fl=1;char ch=gc();
for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
return x*fl;
}
inline void wri(int a){if(a<0)a=-a,putchar('-');if(a>=10)wri(a/10);putchar(a%10|48);}
void add(int x,int y,int z){
e[++tot]=(node){y,h[x],z};
h[x]=tot;
}
void dij(){
memset(dis,63,(n+1)<<2);
dis[s]=0;
q.push((kk){s,0});
while (!q.empty()){
kk u=q.top();q.pop();
if (u.d!=dis[u.u]) continue;//重點
for (int i=h[u.u],v;i;i=e[i].ne){
v=e[i].to;
if (u.d+e[i].w<dis[v]){
dis[v]=u.d+e[i].w;
q.push((kk){v,dis[v]});
}
}
}
}
int main(){
n=rd();m=rd();s=rd();
for (;m--;) x=rd(),y=rd(),z=rd(),add(x,y,z);
dij();
putchar('0');
for (i=2;i<=n;i++) putchar(' '),wri(dis[i]);
}