##題目大意 帶限制的最短路,一個點被一些點所限制,一個點能被到達當且僅當限制它的點都被到達. 你可以釋放無限個機器人代替你跑路.

Solution

記錄$d1$表示到達它的最短路,$d2$為到達所有限制它的點的最短路,那麼實際就是$max(d1,d2)$. 一個點能進行增廣,當且僅當它不被任何點限制. 增廣的時候判一下就OK.

Codes

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

using namespace std;

const int N = 3e3 + 5,M = 7e4 + 5;

struct Edge
{
	int to,next,val;
}edge[M];

struct node
{
	int dis,id;
	int operator < (node b) const
	{
		return dis > b.dis;
	}
};

int n,m,u,v,w,tot;
int st[N],dis[N],dis1[N],dis2[N],du[N],vis[N],V[N][N],cnt[N];
priority_queue <node> q;

void link(int u,int v,int w)
{
	edge[++tot].to = v;
	edge[tot].next = st[u];
	edge[tot].val = w;
	st[u] = tot;
}

void work()
{
	memset(dis1,0x3f,sizeof(dis1));
	dis1[1] = 0;
	q.push((node){0,1});
	while (!q.empty())
	{
		node u = q.top(); q.pop();
		int x = u.id;
		if (vis[x]) continue;
		vis[x] = 1;
		for (int l = st[x] ; l ; l = edge[l].next)
		{
			int v = edge[l].to;
			if (dis1[v] > max(dis1[x],dis2[x]) + edge[l].val) 
			{
				dis1[v] = max(dis1[x],dis2[x]) + edge[l].val;
				if (!du[v]) q.push((node){max(dis1[v],dis2[v]),v});
			}
		}
		for (int j = 1 ; j <= cnt[x] ; j++) 
		{
			int k = V[x][j];
			du[k]--;
			dis2[k] = max(dis2[k],max(dis1[x],dis2[x]));
			if (!du[k]) q.push((node){max(dis1[k],dis2[k]),k});
		}
	}
}

int main()
{
	//freopen("in","r",stdin);
	scanf("%d%d",&n,&m);
	for (int i = 1 ; i <= m ; i++) { scanf("%d%d%d",&u,&v,&w); if (u != v) link(u,v,w); }
	for (int i = 1 ; i <= n ; i++)
	{
		scanf("%d",&du[i]);
		for (int j = 1 ; j <= du[i] ; j++) scanf("%d",&u),V[u][++cnt[u]] = i;
	}
	work();
	printf("%d\n",max(dis1[n],dis2[n]));
	return 0;
}