[题解]CF1321D Navigation System

思路：从终点反向bfs，求出每个点到终点的距离

然后枚举原图的每一条边,计算出度

最后对于路径上每个点判断即可

code

#include<bits/stdc++.h>
namespace my_std
{
	using namespace std;
	#define rep(i,a,b) for (int i=(a);i<=(b);i++)
	#define drep(i,a,b) for (int i=(a);i>=(b);i--)
	#define go(u) for (int i=head[(u)];i;i=e[i].nxt)
	#define pf printf
	#define writeln(x) write(x),putchar('
')
	#define writesp(x) write(x),putchar(' ')
	#define mem(x,v) memset(x,v,sizeof(x))
	typedef long long ll;
	const int INF=0x7fffffff;
	inline int read()
	{
		int sum=0,f=1;
		char c=0;
		while (!isdigit(c))
		{
			if (c=='-') f=-1;
			c=getchar();
		}
		while (isdigit(c))
		{
			sum=(sum<<1)+(sum<<3)+(c^48);
			c=getchar();
		}
		return sum*f;
	}
	void write(int k)
	{
		if (k<0) putchar('-'),k=-k;
		if (k>=10) write(k/10);
		putchar(k%10+'0');
	}
}
using namespace my_std;
const int N=200010;
int n,m,head[N],HEAD[N];
int s,t,cnt1,cnt2;
int dis[N],p[N],out[N];
int maxx=0,minn=0;
struct edge
{
	int to,nxt;
}e[N<<1],E[N<<1];
void add1(int u,int v)
{
	e[++cnt1].to=v;
	e[cnt1].nxt=head[u];
	head[u]=cnt1;
}
void add2(int u,int v)
{
	E[++cnt2].to=v;
	E[cnt2].nxt=HEAD[u];
	HEAD[u]=cnt2;
}
void bfs(int f)
{
	queue<int>q;
	dis[f]=1;
	q.push(f);
	while (!q.empty())
	{
		int u=q.front();
		for (int i=HEAD[u];i;i=E[i].nxt)
		{
			int v=E[i].to;
			if (dis[v]==0)
			{
				dis[v]=dis[u]+1;
				q.push(v);
			}
		}
		q.pop();
	}
}
int main()
{
	n=read(),m=read();
	int u,v;
	rep(i,1,m)
	{
		u=read(),v=read();
		add1(u,v),add2(v,u);
	}
	int size=read();
	rep(i,1,size) p[i]=read();
	bfs(p[size]);
	rep(u,1,n) go(u)
	{
		int v=e[i].to;
		if (dis[v]+1==dis[u])
		{
			out[u]++;
		}
	}
	rep(i,1,size-1)
	{
		if (dis[p[i]]!=dis[p[i+1]]+1) minn++,maxx++;
		else if (dis[p[i]]==dis[p[i+1]]+1&&out[p[i]]>=2) maxx++;
	}
	writesp(minn),writeln(maxx);
}