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
",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）
传送门
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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