PAT (Advanced Level) 1003 dij变形 求最短路的数量

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//dij变形
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int>PII;
const int maxv=500+5;
const int maxe=250000+5;
const int INF=0x3f3f3f3f;
struct Edge{
    int to,next,dist;
};
int cnt=0,head[maxv];
Edge edges[maxe];
void init(int u,int v,int k)
{
    edges[++cnt].next=head[u];
    edges[cnt].to=v;
    edges[cnt].dist=k;
    head[u]=cnt;
}
int v,e,s,en,d[maxv],d1[maxv];
LL res[maxv],num[maxv];
bool done[maxv];
void dij()
{
    memset(d,INF,sizeof(d));
    d[s]=0;
    d1[s]=1;
    priority_queue<PII,vector<PII>,greater<PII> >pq;
    while(!pq.empty()) pq.pop();
    memset(done,false,sizeof(done));
    pq.push(PII(0,s));
    while(!pq.empty())
    {
        PII k=pq.top();
        pq.pop();
        if(done[k.second]) continue;
        if(k.second==en) return;
        done[k.second]=true;
        for(int i=head[k.second];i!=0;i=edges[i].next)
        {
            if(!done[edges[i].to]&&d[k.second]+edges[i].dist<d[edges[i].to]) {
                d[edges[i].to]=k.first+edges[i].dist;
                pq.push(PII(d[edges[i].to],edges[i].to));
                res[edges[i].to]=max(1l*res[edges[i].to],1l*res[k.second]+num[edges[i].to]);
                d1[edges[i].to]=d1[k.second];
            }
            else if(!done[edges[i].to]&&d[k.second]+edges[i].dist==d[edges[i].to]) {
                res[edges[i].to]=max(1l*res[edges[i].to],1l*res[k.second]+num[edges[i].to]);
                d1[edges[i].to]+=d1[k.second];
            }
        }
    }
}
int main()
{
    memset(head,0,sizeof(head));
    scanf("%d%d%d%d",&v,&e,&s,&en);
    s++,en++;
    for(int i=1;i<=v;i++)
    {
        scanf("%lld",&num[i]);
    }
    memcpy(res,num,sizeof(num));
    for(int i=0;i<e;i++)
    {
        int v1,v2,di;
        scanf("%d%d%d",&v1,&v2,&di);
        init(v1+1,v2+1,di);
        init(v2+1,v1+1,di);
    }
    dij();
    printf("%d %lld\n",d1[en],res[en]);
    return 0;
}