FOI冬令营 Day1

打算把省冬的题目放上来，主要是防止自己偷懒不订正

T1、全连（fc）

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Code 

//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\n",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\n",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\n",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\n",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--]='\0';
    }
    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\n",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\n",ans);getans(ans);return 0;
}

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