FCS省选模拟赛 Day5

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Solution

Code 

#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
const int inf=0x3f3f3f3f,MN=105,T=235,S=0,TT=245;
int n,m,k,nx[MN],ny[MN],ans;
bool mp[MN][MN];
int d[TT],q[TT],top;
struct edge{int to,w,nex;}e[MN*MN*2];int hr[TT],cur[TT],en=1;
inline void ins(int f,int t,int w)
{
	e[++en]=(edge){t,w,hr[f]};hr[f]=en;
	e[++en]=(edge){f,0,hr[t]};hr[t]=en;
}
bool bfs(){
	memset(d,0,sizeof d);register int i,j;
	for(d[q[top=i=1]=S]=1;i<=top;i++)
		for(j=hr[q[i]];j;j=e[j].nex)
			if(e[j].w&&!d[e[j].to])
			d[q[++top]=e[j].to]=d[q[i]]+1;
	return d[T];
}
int dfs(int x,int f){
	if(x==T) return f;int used=0;
	for(int &i=cur[x];i;i=e[i].nex)
		if(e[i].w&&d[e[i].to]==d[x]+1){
			int w=dfs(e[i].to,min(e[i].w,f-used));
			used+=w;e[i].w-=w;e[i^1].w+=w;
			if(used==f) return used;
		}
	return d[x]=-1,used;
}
inline void dinic()
{
	while(bfs())
	{
		memcpy(cur,hr,sizeof cur);
		ans-=dfs(S,inf);
	}
}
int main()
{
	register int i,j,x,y;
	n=read();m=read();k=read();
	for(i=1;i<=n;++i) nx[i]=m-read();
	for(i=1;i<=m;++i) ny[i]=n-read();
	for(i=1;i<=k;++i) x=read(),y=read(),--nx[x],--ny[y],mp[x][y]=true;
	for(i=1;i<=n;++i) if(nx[i]<0) return 0*puts("IIllIIll1！");
	for(i=1;i<=m;++i) if(ny[i]<0) return 0*puts("IIllIIll1！");
	for(i=1;i<=n;++i) ins(S,i,nx[i]);
	for(i=1;i<=m;++i) ins(i+MN,T,ny[i]);
	for(i=1;i<=n;++i)for(j=1;j<=m;++j) if(!mp[i][j]) ins(i,j+MN,1);
	ans=n*m-k;dinic();
	return 0*printf("%d
",ans);
}

/*
rongchi : f_n=d_n-a_n a_n:num of n's factor that is not QQn
Sum d_i easy = sum (N/i)
sum a_n =sum mu[i]^2 * (n/i)
mobi : sum mu[i]^2 = sum mu[i]*N/(i^2)
*/
#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
const int MN=35000;
ll mu[MN],prime[MN],tot;
bool mk[MN]; 
inline void init()
{
	mu[1]=1;register int i,j;
	for(i=2;i<MN;++i)
	{
		if(!mk[i]){prime[++tot]=i;mu[i]=-1;}
		for(j=1;j<=tot&&i*prime[j]<MN;++j)
		{
			mk[i*prime[j]]=true;
			if(i%prime[j]==0)
			{
				mu[i*prime[j]]=0;
				break;
			}
			else mu[i*prime[j]]=-mu[i];
		}
	}
}
ll get_d(int N)
{
	ll ans=0;register int l=1,r;
	for(;l<=N;l=r+1)
	{
		r=N/(N/l);
		ans+=1ll*(N/l)*(r-l+1);
	}
	return ans;
}
ll Pre(int N)
{
	register int i;ll ans=0;
	for(i=1;i*i<=N;++i) ans+=1ll*mu[i]*(N/i/i);
	return ans;
}
ll get_a(int N)
{
	register int l=1,r;ll ans=0;
	for(;l<=N;l=r+1)
	{
		r=N/(N/l);
		ans+=1ll*(Pre(r)-Pre(l-1))*(N/l);
	}
	return ans;
}
ll get(int N)
{
	if(!N) return 0ll;
	return get_d(N)-get_a(N);
}
int main()
{
	ll L,R;
	init();L=read(),R=read();
	printf("%lld
",get(R)-get(L-1));
}

/*
后缀自动机+线段树合并
对Fail树进行dfs
每个点的level应该是祖先中满足出现次数大于1的最大lev+1
如果没有，level等于祖先中最大的lev
判断出现次数，用线段树区间求和
为什么实现是每个节点只需考虑它的最大len？
如果有小的len，它每次出现时必然伴随最大的len一同出现，所以不影响
2019/3/20 by pac
*/
#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
const int MN=2e5+5,MX=4e5+5,MS=1e7+5;
char s[MN+5];
int len;
class Seg
{
	int ls[MS],rs[MS],v[MS],root[MX],tot;
	void Md(int &x,int l,int r,int a,int ad)
	{
		if(!x) x=++tot;if(l==r){v[x]+=ad;return;}
		int mid=(l+r)>>1;
		if(a<=mid) Md(ls[x],l,mid,a,ad);
		else Md(rs[x],mid+1,r,a,ad);
		v[x]=v[ls[x]]+v[rs[x]];
	}
	int Qy(int x,int l,int r,int a,int b)
	{
		if(!x) return 0;
		if(l==a&&r==b){return v[x];}
		int mid=(l+r)>>1;
		if(b<=mid) return Qy(ls[x],l,mid,a,b);
		if(a>mid) return Qy(rs[x],mid+1,r,a,b);
		return Qy(ls[x],l,mid,a,mid)+Qy(rs[x],mid+1,r,mid+1,b);
	}
	int Merge(int x,int y,int l,int r)
	{
		if(!x||!y) return x|y;
		int o=++tot;v[o]=v[x]+v[y];
		int mid=(l+r)>>1;
		ls[o]=Merge(ls[x],ls[y],l,mid);
		rs[o]=Merge(rs[x],rs[y],mid+1,r);
		return o;
	}
	public:
		void md(int x,int k){Md(root[x],1,len,k,1);}
		bool qy(int x,int l,int r){return Qy(root[x],1,len,l,r)>=2;}
		int merge(int x,int y){root[x]=Merge(root[x],root[y],1,len);}
}T;
class SAM
{
	int c[MX][26],fa[MX],step[MX],pos[MX];
	int last,cnt,n;
	int ans=0,lev[MX];
	struct ed{int to,nex;}e[MX<<1];int en,hr[MX];
	void ins(int x,int y){e[++en]=(ed){y,hr[x]};hr[x]=en;}
	void pre_dfs(int x)
	{
		register int i;
		for(i=hr[x];i;i=e[i].nex)
			pre_dfs(e[i].to),pos[x]=max(pos[x],pos[e[i].to]),T.merge(x,e[i].to);
	}
	void dfs(int x,int mx)
	{
		if(x!=1&&(T.qy(mx,pos[x]-step[x]+step[mx],pos[x])||step[x]==1)) lev[x]=lev[mx]+1,mx=x;
		else lev[x]=lev[mx];register int i;
		for(i=hr[x];i;i=e[i].nex) dfs(e[i].to,mx);
		ans=max(ans,lev[x]);
	}
	public:
		inline void init()
    	{
        	cnt=last=1;n=0;
        	for(int i=1;i<=n<<1;++i)
            memset(c[i],0,sizeof c[i]),lev[i]=step[i]=fa[i]=0;
    	}
		void Insert(int x)
		{
    		int p=last,np=++cnt;step[np]=step[p]+1;
    		T.md(np,pos[np]=++n);
        	for(;p&&!c[p][x];p=fa[p]) c[p][x]=np;
        	if(!p) fa[np]=1;
        	else 
        	{
        	    int q=c[p][x];
        	    if(step[q]==step[p]+1) fa[np]=q;
        	    else 
        	    {
        	        int nq=++cnt;step[nq]=step[p]+1;
        	        memcpy(c[nq],c[q],sizeof c[q]);
        	        fa[nq]=fa[q];fa[np]=fa[q]=nq;
        	        for(;c[p][x]==q;p=fa[p]) c[p][x]=nq;
        	    }    
        	}
        	last=np;
		}
		void solve()
		{
			register int i;
			for(i=2;i<=cnt;++i) if(fa[i]) ins(fa[i],i);
			pre_dfs(1);dfs(1,1);
			printf("%d
",ans);
		}
}sam;
int main()
{
	scanf("%s",s+1);len=strlen(s+1);sam.init();
	register int i;
	for(i=1;i<=len;++i) sam.Insert(s[i]-'a');
	sam.solve();
	return 0;
}

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