【luogu P3393 逃离僵尸岛】 题解

题目链接：https://www.luogu.org/problemnew/show/P3393
被占领的点可以先连在一个点上然后只需要对这一个点bfs一遍就可以求所有的危险点

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 2 * 1e5 + 10;
ll n, m, k, s, val[2], danger[maxn], ddis[maxn], dis[maxn], charge[maxn], v[maxn], u[maxn];
bool vis[maxn], vvis[maxn];
struct edge{
	ll from, to, next, len;
}e[maxn<<2], ee[maxn<<2];
ll cnt, head[maxn], cntt, headd[maxn];
queue<ll> dq, q;
void add(ll u, ll v, ll w)
{
	e[++cnt].from = u;
	e[cnt].len = w;
	e[cnt].next = head[u];
	e[cnt].to = v;
	head[u] = cnt;
}
void ad(ll u, ll v, ll w)
{
	ee[++cntt].from = u;
	ee[cntt].len = w;
	ee[cntt].next = headd[u];
	ee[cntt].to = v;
	headd[u] = cntt;
}
void SPFA()
{
	while(!q.empty())
	{
		ll now = q.front(); q.pop();
		vis[now] = 0;
		for(ll i = head[now]; i != -1; i = e[i].next)
		{
			if(dis[e[i].to] > dis[now] + e[i].len && danger[e[i].to] == 0)
			{
				dis[e[i].to] = dis[now] + e[i].len;
				if(!vis[e[i].to])
				{
					q.push(e[i].to);
					vis[e[i].to] = 1;
				}
			}
		}
	}
}
void BFS()
{
	while(!dq.empty())
	{
		ll now = dq.front(); dq.pop();
		vis[now] = 0;
		for(ll i = headd[now]; i != -1; i = ee[i].next)
		{	
			if(ddis[ee[i].to] > ddis[now] + ee[i].len)
			{
				ddis[ee[i].to] = ddis[now] + ee[i].len;
				if(!vvis[ee[i].to])
				{
					dq.push(ee[i].to);
					vvis[ee[i].to] = 1;
				}
			}
		}
	}
}
int main()
{
	memset(head, -1, sizeof(head));
	memset(headd, -1, sizeof(headd));
	memset(dis, 127, sizeof(dis));
	memset(ddis, 127, sizeof(ddis));
	scanf("%lld%lld%lld%lld%lld%lld",&n,&m,&k,&s,&val[0],&val[1]);
	for(ll i = 1; i <= k; i++)
	{
		ll u;
		scanf("%lld",&u);
		danger[u] = 1;
		ad(u, 0, 1);
		ad(0, u, 1);
	}
	for(ll i = 1; i <= m; i++)
	{
		scanf("%lld%lld",&u[i],&v[i]);
		ad(u[i], v[i], 1);
		ad(v[i], u[i], 1);
	}
	dq.push(0), ddis[0] = 0, vvis[0] = 1;
	BFS();
	for(ll i = 1; i <= n; i++)
	if(ddis[i] <= s+1)
	charge[i] = 1;
	
	for(ll i = 1; i <= m; i++)
	{
		if(danger[u[i]] != 1 && danger[v[i]] != 1)
		{
			add(u[i], v[i], val[charge[v[i]]]);
			add(v[i], u[i], val[charge[u[i]]]);
		}
	}
	
	q.push(1), dis[1] = 0, vis[1] = 1;
	SPFA();
	
	/*for(ll i = 0; i <= n; i++)
	cout<<charge[i]<<" ";*/

	printf("%lld",dis[n]-val[charge[n]]);
	return 0;
}