【bzoj4199】【Noi2015】品酒大会

• 题解

  • SA+并查集
  • 把ht按大小倒序加入，并查集合并维护答案的变化；
  • SAM
  • 翻转串，求出SAM的parent树就是后缀树，两个串的最长公共后缀是他们lca的len值；
  • 考率一个节点x，那么它子树里的后缀点两两都是len[x]相似的，所以在prent树上做dp即可；
  • 第二问的统计比较麻烦，可以直接写一个后缀树的dfs来统计u的当前儿子和之前的儿子的答案，这样子不用维护次大值；
  • dp的具体方式见bzoj3238

  •  #include<bits/stdc++.h>
     #define fp(i,a,b) for(register int i=a,I=b+1;i<I;++i)
     #define fd(i,a,b) for(register int i=a,I=b-1;i>I;--i)
     #define go(u) for(register int i=fi[u],v=e[i].to;i;v=e[i=e[i].nx].to)
     #define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
     template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,:;}
     template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,:;}
     using namespace std;
     char ss[<<],*A=ss,*B=ss;
     inline char gc(){return A==B&&(B=(A=ss)+fread(ss,,<<,stdin),A==B)?-:*A++;}
     template<class T>inline void sd(T&x){
         char c;T y=;while(c=gc(),(c<||<c)&&c!=-)if(c==)y=-;x=c-;
         while(c=gc(),<c&&c<)x=x*+c-;x*=y;
     }
     inline void gs(char*s){char c;while(c=gc(),c<);*s++=c;while(c=gc(),c>)*s++=c;}
     char sr[<<],z[];int C=-,Z;
     inline void Ot(){fwrite(sr,,C+,stdout),C=-;}
     template<class T>inline void we(T x){
         if(C><<)Ot();if(x<)sr[++C]=,x=-x;
         while(z[++Z]=x%+,x/=);
         while(sr[++C]=z[Z],--Z);sr[++C]=' ';
     }
     const int N=3e5+,M=*N,inf=1e9+;
     typedef long long ll;
     typedef int arr[M];
     int n,w[N];char s[N];
     struct SAM{
         int las,T,ch[M][];arr fa,len,sz;
         SAM(){las=T=;}
         inline void ins(int c,int w){
             int p=las,np;fa[np=las=++T]=,len[np]=len[p]+;
             for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
             mx[T]=mi[T]=w,mx2[T]=-inf,mi2[T]=inf,sz[T]=;
             if(p){
                 int q=ch[p][c],nq;
                 if(len[p]+==len[q])fa[np]=q;
                 else{
                     nq=++T;mx[T]=mx2[T]=-inf,mi[T]=mi2[T]=inf;
                     fa[nq]=fa[q],len[nq]=len[p]+,memcpy(ch[nq],ch[q],*);
                     for(fa[np]=fa[q]=nq;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
                 }
             }
         }
         struct eg{int nx,to;}e[M];
         int ce;arr fi,mx,mx2,mi,mi2,sx;ll sum[M],ans[M];
         inline void add(int u,int v){e[++ce]=(eg){fi[u],v},fi[u]=ce;}
         inline void ck1(int u,int w){if(w>mx[u])mx2[u]=mx[u],mx[u]=w;else if(w>mx2[u])mx2[u]=w;}
         inline void ck2(int u,int w){if(w<mi[u])mi2[u]=mi[u],mi[u]=w;else if(w<mi2[u])mi2[u]=w;}
         void dfs(int u){
             int siz=;
             go(u){
                 dfs(v);siz+=sz[v];
                 ck1(u,mx[v]),ck1(u,mx2[v]);
                 ck2(u,mi[v]),ck2(u,mi2[v]);
    
             }if(siz+sz[u]<)return;
             cmax(ans[len[u]],max((ll)mx[u]*mx2[u],(ll)mi[u]*mi2[u]));
             go(u)sum[len[u]]+=(ll)sz[u]*sz[v],sz[u]+=sz[v];
         }
         inline void sol(){
             mx[]=mx2[]=-inf,mi[]=mi2[]=inf;
             memset(ans,-,sizeof ans);
             fp(i,,T)add(fa[i],i);dfs();
             fd(i,n-,)sum[i]+=sum[i+],cmax(ans[i],ans[i+]);
             fp(i,,n-)we(sum[i]),we(!sum[i]?:ans[i]),sr[++C]='\n';
         }
     }p;
     int main(){
         #ifndef ONLINE_JUDGE
             file("s");
         #endif
         sd(n),gs(s+);fp(i,,n)sd(w[i]);
         fd(i,n,)p.ins(s[i]-'a',w[i]);p.sol();
     return Ot(),;
     }
     //https://kelin.blog.luogu.org/solution-p2178

    推荐luogu大佬的实现

  •  #include<bits/stdc++.h>
     #define inf 0x3f3f3f3f
     #define ll long long
     #define il inline
     using namespace std;
     const int N=;
     int n,lst,w[N],len[N],sz,pa[N],mx0[N],mx1[N],mn0[N],mn1[N],ch[N][],c[N],id[N],cnt[N];
     char s[N];
     ll ans1[N],ans2[N];
     il bool upd1(int x,int y){
         if(y==inf)return false;
         if(mx0[x]==inf||y>mx0[x]){mx1[x]=mx0[x];mx0[x]=y;return true;}
         if(mx1[x]==inf||y>mx1[x]){mx1[x]=y;return false;}
         return false;
     }
     il bool upd2(int x,int y){
         if(y==inf)return false;
         if(mn0[x]==inf||y<mn0[x]){mn1[x]=mn0[x];mn0[x]=y;return true;}
         if(mn1[x]==inf||y<mn1[x]){mn1[x]=y;return false;}
         return false;
     }
     il void ins(int now,int x){
         int p=lst; int np=lst=++sz;
         cnt[np]=;
         mx0[np]=mn0[np]=w[now];
         mx1[np]=mn1[np]=inf;
         len[np]=len[p]+;
         while(p&&!ch[p][x])ch[p][x]=np,p=pa[p];
         if(!p){pa[np]=;return;}
         int q=ch[p][x];
         if(len[q]==len[p]+){pa[np]=q;}
         else {
             int nq=++sz; len[nq]=len[p]+;
             memcpy(ch[nq],ch[q],sizeof(ch[q]));
             pa[nq]=pa[q]; pa[q]=pa[np]=nq;
             while(p&&ch[p][x]==q)ch[p][x]=nq,p=pa[p];
         }
     }
     inline ll max(ll x,ll y){return x>y?x:y;}
     int main(){
         freopen("bzoj4199.in","r",stdin);
         freopen("bzoj4199.out","w",stdout);
         lst=sz=;
         memset(mx0,0x3f,sizeof(mx0));
         memset(mx1,0x3f,sizeof(mx1));
         memset(mn0,0x3f,sizeof(mn0));
         memset(mn1,0x3f,sizeof(mn1));
         scanf("%d%s",&n,s+);
         for(int i=;i<=n>>;i++)swap(s[i],s[n-i+]);
         for(int i=;i<=n;i++)scanf("%d",&w[n-i+]);
         for(int i=;i<=n;i++)ins(i,s[i]-'a');
         for(int i=;i<=sz;i++)c[len[i]]++;
         for(int i=;i<=n;i++)c[i]+=c[i-];
         for(int i=sz;i;i--)id[c[len[i]]--]=i,ans2[i]=-1e18;
         len[]=-;
         for(int i=sz;i;i--){
             int u=id[i];
             ll t1=(ll)cnt[u]*(cnt[u]-)/;
             ll t2=max((ll)mx0[u]*mx1[u], (ll)mn0[u]*mn1[u]);
             ans1[len[u]]+=t1;
             if(t1)ans2[len[u]]=max(ans2[len[u]],t2);
             ans1[len[pa[u]]]-=t1;
             cnt[pa[u]]+=cnt[u];
             if(upd1(pa[u],mx0[u]))upd1(pa[u],mx1[u]);
             if(upd2(pa[u],mn0[u]))upd2(pa[u],mn1[u]);
         }
         for(int i=n-;~i;i--){
             ans1[i]+=ans1[i+];
             ans2[i]=max(ans2[i],ans2[i+]);
         }
         for(int i=;i<n;i++){
             if(ans1[i])printf("%lld %lld\n",ans1[i],ans2[i]);
             else puts("0 0");
         }
         return ;
     }

    bzoj4199