dij預處理水廠到各個點的最短路，對於每條邊，我們做一下處理
單向邊
ans=max(ans,f[i]+len[j])
雙向邊有可能是兩邊合起來，所以
Ans−f[x]+Ans−f[y]>=len[j]
d=len[j]+f[x]+f[y]
2Ans=d
ans=max(ans,d/2+d%2)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;
int
 
 n,m,x,y,z,t,ty,tot;
int a[25005],heap[25005],pos[25005],head[25005];
ll f[25005];
int Next[600005],to[600005],len[600005],flag[600005];
void up(int id)
{
    while(id/2>=1) 
    {
        if(f[heap[id]]<f[heap[id/2]]) 
        {
            swap(heap[id],heap[id/2]);
            swap(pos[heap[id]],pos[heap[id/2]]);
            id=id/2
 
;
        }
        else break;
    }
}
void down(int id)
{
    while(id*2<=ty) 
    {
        int j;
        if(id*2+1>ty||f[heap[id*2+1]]>f[heap[id*2]]) j=id*2;else j=id*2+1;
        if(f[heap[j]]<f[heap[id]])
        {
            swap(heap[j],heap[id]);
            swap(pos[heap[j]],pos
 
[heap[id]]);
            id=j;
        }
        else break;
    }
}
ll dij() 
{
    for(int i=1;i<=n;i++) 
    if(a[i]==1) 
    {
        ty++;
        heap[ty]=i;
        pos[i]=ty;
        f[i]=0;
    }
    else
    {
        pos[i]=0;
        f[i]=1e10;
    }
    while(ty>0) 
    {
        int x=heap[1];
        swap(heap[1],heap[ty]);
        swap(pos[heap[1]],pos[heap[ty]]);
        ty--;
        down(1);
        for(int i=head[x];i!=-1;i=Next[i]) 
        if(f[x]+len[i]<f[to[i]]) 
        {
            f[to[i]]=f[x]+len[i];
            if(pos[to[i]]==0) 
            {
                ty++;
                heap[ty]=to[i];
                pos[to[i]]=ty;
            }
            up(pos[to[i]]);
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++) 
    for(int j=head[i];j!=-1;j=Next[j]) 
    if(flag[j]==1) 
    {
        ans=max(ans,f[i]+len[j]);
    }
    else
    {
        ll dis=f[i]+f[to[j]]+len[j];
        ans=max(ans,dis/2+dis%2);
    }
    return ans;
}
void add(int x,int y,int z,int t)
{
    Next[tot]=head[x];
    to[tot]=y;
    len[tot]=z;
    flag[tot]=t;
    head[x]=tot;
    tot++;
}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=n;i++) 
    {
        scanf("%d",&a[i]);
        head[i]=-1;
    }
    for(int i=1;i<=m;i++) 
    {
        scanf("%d%d%d%d",&x,&y,&t,&z);
        if(t==1) 
        {
            add(x,y,z,1);
        }
        else
        {
            add(x,y,z,2);
            add(y,x,z,2);
        }
    }
    cout<<dij();
    return 0;
}