【JZOJ6411】上网

description

---

analysis

• 如果把所有大小关系连成边，小的往大的连，就可以直接上拓扑

• 暴力连边时间复杂度\(O(n^2)\)，然而连边的过程，考虑用线段树优化

• 线段树上的所有儿子节点向父亲节点连\(0\)边，

• 每个操作被分成\(k+1\)个小区间，然后该操作的编号向\(k\)个区间最大值连\(1\)边

• 对于线段树上表示小区间的\(\log\)个区间，都向该编号连\(0\)边

• 最后一波拓扑就好了，这类套路要记一下

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#define MAXN 100005
#define ll long long
#define reg register ll
#define max(x,y) ((x>y)?(x):(y))
#define min(x,y) ((x<y)?(x):(y))
#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 rep(i,a) for (reg i=last[a];i;i=next[i])

using namespace std;

ll last[MAXN*50],next[MAXN*50],tov[MAXN*50],len[MAXN*50];
ll a[MAXN*20],d[MAXN*20],pos[MAXN];
ll n,m,q,now,tot;
queue<ll>Q;

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;
}
inline void link(ll x,ll y,ll z){next[++tot]=last[x],last[x]=tot,tov[tot]=y,len[tot]=z,++d[y];}
inline void maketree(ll t,ll l,ll r)
{
	now=max(now,t);if (l==r){pos[l]=t;return;}
	ll mid=(l+r)>>1;link(t<<1,t,0),link((t<<1)+1,t,0);
	maketree(t<<1,l,mid),maketree((t<<1)+1,mid+1,r);
}
inline void query(ll t,ll l,ll r,ll x,ll y)
{
	if (l==x && y==r){link(t,now,0);return;}
	ll mid=(l+r)>>1;
	if (x<=mid)query(t<<1,l,mid,x,min(y,mid));
	if (y>mid)query((t<<1)+1,mid+1,r,max(x,mid+1),y);
}
int main()
{
	freopen("web.in","r",stdin);
	//freopen("web.out","w",stdout);
	n=read(),m=read(),q=read();
	maketree(1,1,n);
	fo(i,1,m){ll x=read();a[pos[x]]=read();}
	while (q--)
	{
		ll l=read(),r=read(),k=read(),x=l;++now;
		fo(i,1,k)
		{
			ll y=read();if (x<=y-1)query(1,1,n,x,y-1);
			link(now,pos[y],1),x=y+1;
		}
		if (x<=r)query(1,1,n,x,r);
	}
	fo(i,1,now)if (!d[i]){Q.push(i);if (!a[i])a[i]=1;}
	while (!Q.empty())
	{
		ll x=Q.front();Q.pop();
		rep(i,x)
		{
			a[tov[i]]=max(a[tov[i]],a[x]+len[i]);
			if (!(--d[tov[i]]))Q.push(tov[i]);
		}
	}
	printf("Possible\n");
	fo(i,1,n)printf("%lld ",a[pos[i]]);
	printf("\n");
	return 0;
}