【JZOJ3155】最短路

description

N 个结点、M 个含K 个结点的完全子图构成一个奇怪的图，问从结点1 走
到结点N 最少需要经过多少个结点。

---

analysis

• $bfs$

• 对于一个小完全图，不要暴力两两连边

• 其实我们把它连成一个菊花图，用中间的点连向这些点即可

• 然后跑一次宽搜

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#define MAXN 110005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define O3 __attribute__((optimize("-O3")))

using namespace std;

vector<ll>v[MAXN];
ll n,m,k,ans=-1;
bool bz[MAXN];
ll val[MAXN];

struct node
{
	ll x,dis;
}que[MAXN];

O3 inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
O3 inline ll max(ll x,ll y)
{
	return x>y?x:y;
}
O3 int main()
{
	//freopen("T2.in","r",stdin);
	n=read(),k=read(),m=read();
	fo(i,1,n)val[i]=1;
	fo(i,1,m)
	{
		val[n+i]=0;
		fo(j,1,k)
		{
			ll x=read();
			v[n+i].push_back(x),v[x].push_back(n+i);
		}
	}
	memset(bz,1,sizeof(bz)),bz[1]=0;
	ll l=0,r=1;que[1].x=1,que[1].dis=1;
	while (l<r)
	{
		node now=que[++l];
		if (now.x==n)
		{
			ans=now.dis;
			break;
		}
		if (!v[now.x].size())continue;
		fo(i,0,v[now.x].size()-1)if (bz[v[now.x][i]])
		{
			bz[v[now.x][i]]=0;
			que[++r].x=v[now.x][i],que[r].dis=now.dis+val[v[now.x][i]];
		}
	}
	printf("%lld\n",ans);
	return 0;
}