BZOJ 1922 大陆争霸

神奇Dij
不是很懂
%hzwer代码

蒟蒻我写了个\(N^2\)的TLE，也不是正确性怎样TAT

#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

int read(){
    int x=0, f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0' && ch<='9'){x=x*10+(ch-'0');ch=getchar();}
    return x*f;
}

const int MAXN=3011;
const int MAXM=70111;
const int INF=1034567890;

int N, M;

struct Vert{
    int FE;
    int Deg;
    int Dis, Wait;
    bool Vis;
} V[MAXN];

struct Edge{
    int x, y, next;
    int l;
} E[MAXM];

int Ecnt=0;

void addE(int a, int b, int c){
    ++Ecnt;
    E[Ecnt].x=a;E[Ecnt].y=b;E[Ecnt].l=c;
    E[Ecnt].next=V[a].FE;V[a].FE=Ecnt;
}

int P[MAXN][MAXN], Pcnt[MAXN];

priority_queue< pair<int, int>, vector< pair<int, int> >, greater< pair<int, int> > > PQ;

void Dij(){
    
    for(int i=1;i<=N;++i)
        V[i].Dis=INF;
    
    PQ.push(make_pair(0, 1));
    V[1].Dis=0;
    
    int at, di;
    while(!PQ.empty()){
        at=PQ.top().second;PQ.pop();
        if(V[at].Vis)   continue;
        V[at].Vis=true;
        di=max(V[at].Dis, V[at].Wait);
        for(int k=V[at].FE, to;k>0;k=E[k].next){
            to=E[k].y;
            if(di+E[k].l<V[to].Dis){
                V[to].Dis=di+E[k].l;
                if(V[to].Deg==0)
                    PQ.push(make_pair(max(V[to].Dis, V[to].Wait), to));
            }
        }
        for(int i=1, to;i<=Pcnt[at];++i){
            to=P[at][i];
            --V[to].Deg;V[to].Wait=max(V[to].Wait, di);
            if(V[to].Deg==0)
                PQ.push(make_pair(max(V[to].Wait, V[to].Dis), to));
        }
    }
}

int main(){
    
    N=read();M=read();
    for(int i=1, a, b, c;i<=M;++i){
        a=read();b=read();c=read();
        if(a==b)    continue;/**/
        addE(a, b, c);
    }
    for(int i=1, c;i<=N;++i){
        c=read();
        V[i].Deg=c;
        for(int j=1, p;j<=c;++j){
            p=read();
            P[p][++Pcnt[p]]=i;
        }
    }
    Dij();
    
    printf("%d\n", max(V[N].Dis, V[N].Wait));
    
    return 0;
}

/*
6 6
1 2 1
1 4 3
2 3 1
2 5 2
4 6 2
5 3 2
0
0
0
1 3
0
2 3 5

5

*/