T1、全连（fc）

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//PaperCloud 2019/2/12
//60 pts
#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()
{
	register int x=0,f=1;register char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 1000005
int N,tm[MN];
ll a[MN];
namespace solve1
{
	ll f[10005],ans;
	void work()
	{
		register int i,j;
		memset(f,0,sizeof f);
		ans=f[1]=a[1];
		for(i=2;i<=N;++i)
		{
			f[i]=a[i]; 
			for(j=i-1;j;--j) if(max(tm[i],tm[j])<=i-j) f[i]=max(f[i],f[j]+a[i]);
			ans=max(ans,f[i]);
		}
		printf("%lld
",ans);
	}
}
namespace solve2
{
	struct Node
	{
		ll val,lazy;
	}T[MN<<3];
	void down(int x)
	{
		if(!T[x].lazy) return;
		T[x<<1].val=max(T[x<<1].val,T[x].lazy);
		T[x<<1].lazy=max(T[x<<1].lazy,T[x].lazy);
		T[x<<1|1].val=max(T[x<<1|1].val,T[x].lazy);
		T[x<<1|1].lazy=max(T[x<<1|1].lazy,T[x].lazy);
		T[x].lazy=0;
	}
	void Mdf(int x,int l,int r,int a,int b,ll val)
	{
		if(a==l&&r==b) {T[x].val=max(T[x].val,val);T[x].lazy=max(T[x].lazy,val);return;}
		register int mid=(l+r)>>1;down(x);
		if(b<=mid) Mdf(x<<1,l,mid,a,b,val);
		else if(a>mid) Mdf(x<<1|1,mid+1,r,a,b,val);
		else Mdf(x<<1,l,mid,a,mid,val),Mdf(x<<1|1,mid+1,r,mid+1,b,val);
		T[x].val=max(T[x<<1].val,T[x<<1|1].val);
	}
	ll Gi(int x,int l,int r,int p)
	{
		if(l==r) return T[x].val;
		register int mid=(l+r)>>1;down(x);
		if(p<=mid) return Gi(x<<1,l,mid,p);
		else return Gi(x<<1|1,mid+1,r,p);
	}
	void work()
	{
		register int i;
		Mdf(1,1,N<<1,tm[1]+1,N<<1,a[1]);
		for(i=2;i<=N;++i)
		{
			//printf("update: %d 10 %lld",i+tm[i],a[i]+Gi(1,1,N<<1,i));
			Mdf(1,1,N<<1,i+tm[i],N<<1,a[i]+Gi(1,1,N<<1,i));
		}
		printf("%lld
",T[1].val);
	}
}
int main()
{
	freopen("fc.in","r",stdin);
	freopen("fc.out","w",stdout);
	N=read();
	register int i,j;
	for(i=1;i<=N;++i) tm[i]=read();
	for(i=1;i<=N;++i) a[i]=1ll*read()*tm[i];
	bool flag=1;
	for(i=1;i<=N;++i) if(tm[1]!=tm[i]) {flag=0;break;}
	if(flag==1||N>10000) solve2::work();	
	else if(N<=10000) solve1::work();
	return 0;
}

/*
	首先 写出一个dp
	按照 i+tm[i] 从小到大考虑
	维护 前缀的最大值
	2019/2/12 19:40~20:02
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline ll read()
{
	ll 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;
}
#define MN 1000005 
ll a[MN],ti[MN],t[MN],n,ans,f[MN];
inline void rw(ll &x,ll y){if(y>x)x=y;}
inline void C(int p,ll val){for(;p<=n;p+=(p&(-p))) rw(t[p],val);}
inline ll G(int p){ll r=0;for(;p>0;p-=(p&(-p))) rw(r,t[p]);return r;}
std::vector<int> g[MN];
int main()
{
	freopen("fc.in","r",stdin);
	freopen("fc.out","w",stdout);
	n=read();
	register int i;
	for(i=1;i<=n;++i) ti[i]=read();
	for(i=1;i<=n;++i) a[i]=1ll*read()*ti[i];
	for(i=1;i<=n;++i)
	{
		if(i+ti[i]<=n) g[i+ti[i]].push_back(i);
		for(int j=g[i].size()-1;~j;--j) C(g[i][j],f[g[i][j]]);
		f[i]=G(i-ti[i])+a[i];
		rw(ans,f[i]); 
	}
	return 0*printf("%lld
",ans);
}

T2、原样输出（copy）

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Code

//PaperCloud 2019/2/12
//40 pts
#include<bits/stdc++.h>
#define ll long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define MX 1048580
#define mod 1000000007
#define MM MX<<1
int c[MM][5],fa[MM],step[MM],val[MM];
int v[MM],rk[MM],siz[MM];
int last=1,cnt=1,Line,K,ans,n;
void Insert(int x)
{
	int p=last,np=++cnt;step[np]=step[p]+1;val[np]=1;
    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;
}
inline int Num(char x)
{
	if(x=='A') return 1;
	if(x=='C') return 2;
	if(x=='G') return 3;
	if(x=='T') return 4;
}
inline void putch(int x)
{
	if(x==1) putchar('A');
	if(x==2) putchar('C');
	if(x==3) putchar('G');
	if(x==4) putchar('T');
}
namespace solve1
{
	char str[MX];int st[MX],nn;
	inline void dfs2(int x)
	{
		register int i;
		ans++;ans%=mod;
		for(int j=1;j<=nn;++j) putch(st[j]);puts("");
		for(i=1;i<=4;++i) if(c[x][i])
		{
			st[++nn]=i;
			dfs2(c[x][i]);
			--nn;
		}
	}
	inline void ddd()
	{
		register int i,j;
		for(i=1;i<=cnt;++i) ++v[step[i]];
    	for(i=1;i<=n;++i) v[i]+=v[i-1];
    	for(i=1;i<=cnt;++i) rk[v[step[i]]--]=i;
		for(i=cnt;i;--i) siz[i]=1;
		for(i=cnt;i;--i)for(j=1;j<5;++j)if(c[rk[i]][j]) (siz[rk[i]]+=siz[c[rk[i]][j]])%=mod;
		ans=siz[1]%mod;
	}
	void work()
	{
		register int i;
		scanf("%s",str+1);n=strlen(str+1);
		for(i=1;i<=n;++i) Insert(Num(str[i]));
		scanf("%d",&K);
		if(K==0) ddd();else dfs2(1);
		printf("%d
",ans);
	}
}
int main()
{
	freopen("copy.in","r",stdin);
	freopen("copy.out","w",stdout);
	scanf("%d",&Line);
	if(Line==1) solve1::work();
	//else if(Line<3) solve2::work();
	return 0;
}

/*
	考虑每个串建一个后缀自动机 并且把他们连在一起
	具体的 如果在当前点失配 就找到下一个包含失配字符的自动机
	这样采用的时贪心的思想，每个字符串所对应的状态仍然是唯一的 
	学习一下读入优化？ 
	2019/2/12 20:26~21:32
*/
#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
namespace IO
{
	const int lim=(1<<20)+5;
	char buf[lim+5],*S,*T;
	inline char gc(){if(S==T){T=(S=buf)+fread(buf,1,lim,stdin);if(S==T)return EOF;}return *S++;}
	inline int read()
	{
		int x;char ch;bool f;
		for(f=0;(ch=gc())<'0'||ch>'9';f=ch=='-');
		for(x=ch^'0';(ch=gc())>='0'&&ch<='9';x=(x<<1)+(x<<3)+(ch^'0'));
		return f?-x:x;
	}
	inline int Num(char x)
	{
		if(x=='A') return 0;
		else if(x=='C') return 1;
		else if(x=='G') return 2;
		else if(x=='T') return 3;
		else return -1;
	}
}
using namespace IO; 
const char alpha[4]={'A','C','G','T'};
int n,ans,k;
class Suf_Automation
{
	private:
	#define MN 1048580
	int ch[MN<<1][4],cnt,last,step[MN<<1],sz[MN<<1],fa[MN<<1],rt[MN];
	inline void Insert(int R,int x)
	{
		int np=++cnt,p;step[np]=step[last]+1;
		for(p=last;!ch[p][x];p=fa[p]) ch[p][x]=np;
		if(!p) fa[np]=R;
		else
		{
			int q=ch[p][x];
			if(step[q]==step[p]+1) fa[np]=q;
			else
			{
				int nq=++cnt;step[nq]=step[p]+1;
				memcpy(ch[nq],ch[q],sizeof ch[nq]);
				fa[nq]=fa[q];fa[q]=fa[np]=nq;
				for(;ch[p][x]==q;p=fa[p]) ch[p][x]=nq;
			}
		}
		last=np;
	}
	inline void ins(int i)
	{
		register char c;
		while(!(~Num(c=gc())));
		last=rt[i]=++cnt;
		for(;~Num(c);c=gc()) Insert(rt[i],Num(c));
	}
	char st[MN];int tp;
	inline void dfs(int x)
	{
		if(!x) return ;
		++ans;puts(st+1);
		for(int al=0;al<4;++al) st[++tp]=alpha[al],dfs(ch[x][al]),st[tp--]='';
	}
	int Dfs(int x)
	{
		if(!x) return 0;
		if(sz[x]) return sz[x];
		sz[x]=1;
		for(int al=0;al<4;++al) (sz[x]+=Dfs(ch[x][al]))%=mod;
		return sz[x];
	}
	public:
	void solve()
	{
		n=read();register int i,j,al;
		for(i=1;i<=n;++i) ins(i);
		k=read();
		for(i=n-1;i;--i)for(j=rt[i+1]-1;j>=rt[i];--j)for(al=0;al<4;++al)
			if(!ch[j][al]) ch[j][al]=ch[rt[i+1]][al];
		if(k==1) dfs(1);
		else ans=Dfs(1);
		printf("%d
",ans);
	}
}pac;
int main()
{
	freopen("copy.in","r",stdin);
	freopen("copy.out","w",stdout);
	pac.solve();
	return 0;
}

T3、不同的缩写（diff）

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Code

/*
	每个字符串只需要找出n个子序列即可（越短越好）
	二分答案，dinic跑匹配
	2019/2/13 12:07~13:00+15:30~16:07
*/ 
#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=305,MS=305,TT=100505;
int N,ch[MN][MS][26],len[MN];
char s[MN][MS];
int trie[MN*MN+10][26],tot,siz[MN*MN+10];
char t[MN*MN][MS];
int num;
std::vector<int> G[MN];
std::queue<std::pair<int,int> > que;
class Dinic
{
	private:
		const int S=0,T=100500,inf=0x3f3f3f3f;
		struct edge{int to,w,nex;}e[TT*2];int cur[TT],hr[TT],d[TT],q[TT],en;
		inline void Ins(int f,int t)
		{
			e[++en]=(edge){t,1,hr[f]};hr[f]=en;
			e[++en]=(edge){f,0,hr[t]};hr[t]=en;
		}
		inline bool bfs()
		{
			memset(d,0,sizeof d);register int i,j,tp;
			for(d[q[i=tp=1]=S]=1;i<=tp;++i)
			for(j=hr[q[i]];j;j=e[j].nex)
				if(!d[e[j].to]&&e[j].w) d[q[++tp]=e[j].to]=d[q[i]]+1;
			return d[T];
		}
		inline int dfs(int x,int f)
		{
			if(x==T) return f;register int used=0;
			for(int &i=cur[x];i;i=e[i].nex)
			if(d[e[i].to]==d[x]+1&&e[i].w)
			{
				int w=dfs(e[i].to,min(f-used,e[i].w));
				used+=w;e[i].w-=w;e[i^1].w+=w;
				if(used==f) return used;
			}
			return d[x]=-1,used;
		}
	public:
		inline void ins(int f,int t){Ins(f,t+N);}
		inline void init(){memset(hr,0,sizeof hr);en=1;}
		bool check()
		{
			for(int i=1;i<=N;++i) Ins(S,i);
			for(int i=1;i<=tot;++i) Ins(i+N,T);
			int maxflow=0;
			while(bfs()) memcpy(cur,hr,sizeof cur),maxflow+=dfs(S,inf);
			return maxflow==N;
		}
		inline void G()
		{
			for(int i=1;i<=N;++i) for(int j=hr[i];j;j=e[j].nex)
			if(!e[j].w&&e[j].to) {puts(t[e[j].to-N]);break;}
		}
}pac;
bool chk(int mid)
{
	register int i,j,S;pac.init();
	for(i=1;i<=N;++i) for(S=G[i].size(),j=0;j<S;++j)
		if(siz[G[i][j]]<=mid) pac.ins(i,G[i][j]);
	return pac.check();
}
inline void getans(int ans){chk(ans);pac.G();}
void bfs(int str)
{
	while(!que.empty()) que.pop();
	que.push(std::make_pair(0,0));
	while(!que.empty())
	{
		int last=que.front().first,p=que.front().second;que.pop();
		register int i;
		for(i=0;i<26;++i)if(ch[str][p][i]<=len[str])
		{
			if(!trie[last][i])
			{
				trie[last][i]=++tot;
				siz[tot]=siz[last]+1;
				for(int al=0;al<siz[last];++al) t[tot][al]=t[last][al];
				t[tot][siz[last]]='a'+i;	
			}
			++num;G[str].push_back(trie[last][i]);
			que.push(std::make_pair(trie[last][i],ch[str][p][i]));
			if(num>=N) return;
		}
	}
}
int main()
{
	freopen("diff.in","r",stdin);
	freopen("diff.out","w",stdout);
	scanf("%d",&N);
	register int i,j;
	for(i=1;i<=N;++i) scanf("%s",s[i]+1),len[i]=strlen(s[i]+1);
	for(i=1;i<=N;++i)
	{
		memset(ch[i][len[i]],0x3f,sizeof ch[i][len[i]]);
		for(j=len[i]-1;~j;--j)
			memcpy(ch[i][j],ch[i][j+1],sizeof ch[i][j]),
			ch[i][j][s[i][j+1]-'a']=j+1;
	}
	for(i=1;i<=N;++i) num=0,bfs(i);
	int mid,l=1,r=300,ans=-1;
	for(;l<=r;mid=(l+r)>>1,chk(mid)?(ans=mid,r=mid-1):(l=mid+1));
	if(ans==-1) return 0*puts("-1");
	printf("%d
",ans);getans(ans);return 0;
}

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