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 1048580int 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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