Continental hegemony "SDOI2010"

Ideas:

\ (Robot by the time a city = max (a city guard time to lift the robot arrival time here) \)

Code:

#include<bits/stdc++.h>

using namespace std;

struct E
{
    int to, next, val;
};
E edge[70005];
int n, m, tot, first[3005], a[3005][3005], pro[3005], p[3005]; //pro:每个城市保护多少城市   p:每个城市受多少城市保护
void addedge(int x, int y, int z)
{
    tot++;
    edge[tot].to = y;
    edge[tot].next = first[x];
    edge[tot].val = z;
    first[x] = tot;
}
priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > heap;
bool vis[3005];
int d1[3005], d2[3005], d[3005]; //d1:到达时间 d2:可进入时间
void dijkstra()
{
    memset(d, 63, sizeof(d));
    memset(d1, 63, sizeof(d1));
    heap.push(make_pair(0, 1));
    d1[1] = 0;
    d[1] = 0;
    while(!heap.empty())
    {
        int now = heap.top().second;
        heap.pop();
        if(vis[now])
        {
            continue;
        }
        vis[now] = true;
        for(int u = first[now]; u; u = edge[u].next)
        {
            int v = edge[u].to;
            if(d[now] + edge[u].val < d1[v]) //到达时间
            {
                d1[v] = d[now] + edge[u].val;
                d[v] = max(d1[v], d2[v]);
                if(!p[v])
                {
                    heap.push(make_pair(d[v], v));
                }
            }
        }
        for(int i = 1; i <= pro[now]; i++) //更新保护城市
        {
            int v = a[now][i];
            p[v]--;
            d2[v] = max(d2[v], d[now]);
            d[v] = max(d1[v], d2[v]);
            if(!p[v])
            {
                heap.push(make_pair(d[v], v));
            }
        }
    }
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> n >> m;
    for(int i = 1; i <= m; i++)
    {
        int u, v, w;
        cin >> u >> v >> w;
        addedge(u, v, w);
    }
    for(int i = 1; i <= n; i++)
    {
        cin >> p[i];
        for(int j = 1, x; j <= p[i]; j++)
        {
            cin >> x;
            pro[x]++;
            a[x][pro[x]] = i;
        }
    }
    dijkstra();
    cout << d[n] << endl;
    return 0;
}