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

 1 #include<bits/stdc++.h>
 2 #define fp(i,a,b) for(register int i=a,I=b+1;i<I;++i)
 3 #define fd(i,a,b) for(register int i=a,I=b-1;i>I;--i)
 4 #define go(u) for(register int i=fi[u],v=e[i].to;i;v=e[i=e[i].nx].to)
 5 #define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
 6 template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
 7 template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
 8 using namespace std;
 9 char ss[1<<17],*A=ss,*B=ss;
10 inline char gc(){return A==B&&(B=(A=ss)+fread(ss,1,1<<17,stdin),A==B)?-1:*A++;}
11 template<class T>inline void sd(T&x){
12     char c;T y=1;while(c=gc(),(c<48||57<c)&&c!=-1)if(c==45)y=-1;x=c-48;
13     while(c=gc(),47<c&&c<58)x=x*10+c-48;x*=y;
14 }
15 inline void gs(char*s){char c;while(c=gc(),c<32);*s++=c;while(c=gc(),c>32)*s++=c;}
16 char sr[1<<21],z[20];int C=-1,Z;
17 inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
18 template<class T>inline void we(T x){
19     if(C>1<<20)Ot();if(x<0)sr[++C]=45,x=-x;
20     while(z[++Z]=x%10+48,x/=10);
21     while(sr[++C]=z[Z],--Z);sr[++C]=' ';
22 }
23 const int N=3e5+5,M=2*N,inf=1e9+7;
24 typedef long long ll;
25 typedef int arr[M];
26 int n,w[N];char s[N];
27 struct SAM{
28     int las,T,ch[M][26];arr fa,len,sz;
29     SAM(){las=T=1;}
30     inline void ins(int c,int w){
31         int p=las,np;fa[np=las=++T]=1,len[np]=len[p]+1;
32         for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
33         mx[T]=mi[T]=w,mx2[T]=-inf,mi2[T]=inf,sz[T]=1;
34         if(p){
35             int q=ch[p][c],nq;
36             if(len[p]+1==len[q])fa[np]=q;
37             else{
38                 nq=++T;mx[T]=mx2[T]=-inf,mi[T]=mi2[T]=inf;
39                 fa[nq]=fa[q],len[nq]=len[p]+1,memcpy(ch[nq],ch[q],4*26);
40                 for(fa[np]=fa[q]=nq;ch[p][c]==q;p=fa[p])ch[p][c]=nq;
41             }
42         }
43     }
44     struct eg{int nx,to;}e[M];
45     int ce;arr fi,mx,mx2,mi,mi2,sx;ll sum[M],ans[M];
46     inline void add(int u,int v){e[++ce]=(eg){fi[u],v},fi[u]=ce;}
47     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;}
48     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;}
49     void dfs(int u){
50         int siz=0;
51         go(u){
52             dfs(v);siz+=sz[v];
53             ck1(u,mx[v]),ck1(u,mx2[v]);
54             ck2(u,mi[v]),ck2(u,mi2[v]);
55 
56         }if(siz+sz[u]<2)return;
57         cmax(ans[len[u]],max((ll)mx[u]*mx2[u],(ll)mi[u]*mi2[u]));
58         go(u)sum[len[u]]+=(ll)sz[u]*sz[v],sz[u]+=sz[v];
59     }
60     inline void sol(){
61         mx[1]=mx2[1]=-inf,mi[1]=mi2[1]=inf;
62         memset(ans,-63,sizeof ans);
63         fp(i,2,T)add(fa[i],i);dfs(1);
64         fd(i,n-1,0)sum[i]+=sum[i+1],cmax(ans[i],ans[i+1]);
65         fp(i,0,n-1)we(sum[i]),we(!sum[i]?0:ans[i]),sr[++C]='\n';
66     }
67 }p;
68 int main(){
69     #ifndef ONLINE_JUDGE
70         file("s");
71     #endif
72     sd(n),gs(s+1);fp(i,1,n)sd(w[i]);
73     fd(i,n,1)p.ins(s[i]-'a',w[i]);p.sol();
74 return Ot(),0;
75 }
76 //https://kelin.blog.luogu.org/solution-p2178

推荐luogu大佬的实现

 1 #include<bits/stdc++.h>
 2 #define inf 0x3f3f3f3f
 3 #define ll long long
 4 #define il inline 
 5 using namespace std;
 6 const int N=600010;
 7 int n,lst,w[N],len[N],sz,pa[N],mx0[N],mx1[N],mn0[N],mn1[N],ch[N][26],c[N],id[N],cnt[N];
 8 char s[N];
 9 ll ans1[N],ans2[N];
10 il bool upd1(int x,int y){
11     if(y==inf)return false;
12     if(mx0[x]==inf||y>mx0[x]){mx1[x]=mx0[x];mx0[x]=y;return true;}
13     if(mx1[x]==inf||y>mx1[x]){mx1[x]=y;return false;}
14     return false;
15 }
16 il bool upd2(int x,int y){
17     if(y==inf)return false;
18     if(mn0[x]==inf||y<mn0[x]){mn1[x]=mn0[x];mn0[x]=y;return true;}
19     if(mn1[x]==inf||y<mn1[x]){mn1[x]=y;return false;}
20     return false;
21 } 
22 il void ins(int now,int x){
23     int p=lst; int np=lst=++sz;
24     cnt[np]=1;
25     mx0[np]=mn0[np]=w[now];
26     mx1[np]=mn1[np]=inf;
27     len[np]=len[p]+1;
28     while(p&&!ch[p][x])ch[p][x]=np,p=pa[p];
29     if(!p){pa[np]=1;return;}
30     int q=ch[p][x];
31     if(len[q]==len[p]+1){pa[np]=q;}
32     else {
33         int nq=++sz; len[nq]=len[p]+1;
34         memcpy(ch[nq],ch[q],sizeof(ch[q])); 
35         pa[nq]=pa[q]; pa[q]=pa[np]=nq;
36         while(p&&ch[p][x]==q)ch[p][x]=nq,p=pa[p]; 
37     }
38 }
39 inline ll max(ll x,ll y){return x>y?x:y;}
40 int main(){
41     freopen("bzoj4199.in","r",stdin);
42     freopen("bzoj4199.out","w",stdout);
43     lst=sz=1; 
44     memset(mx0,0x3f,sizeof(mx0));
45     memset(mx1,0x3f,sizeof(mx1));
46     memset(mn0,0x3f,sizeof(mn0));
47     memset(mn1,0x3f,sizeof(mn1));
48     scanf("%d%s",&n,s+1);
49     for(int i=1;i<=n>>1;i++)swap(s[i],s[n-i+1]);
50     for(int i=1;i<=n;i++)scanf("%d",&w[n-i+1]);
51     for(int i=1;i<=n;i++)ins(i,s[i]-'a');
52     for(int i=1;i<=sz;i++)c[len[i]]++;
53     for(int i=1;i<=n;i++)c[i]+=c[i-1]; 
54     for(int i=sz;i;i--)id[c[len[i]]--]=i,ans2[i]=-1e18;
55     len[0]=-1;
56     for(int i=sz;i;i--){
57         int u=id[i];
58         ll t1=(ll)cnt[u]*(cnt[u]-1)/2;
59         ll t2=max((ll)mx0[u]*mx1[u], (ll)mn0[u]*mn1[u]); 
60         ans1[len[u]]+=t1; 
61         if(t1)ans2[len[u]]=max(ans2[len[u]],t2);
62         ans1[len[pa[u]]]-=t1;
63         cnt[pa[u]]+=cnt[u];
64         if(upd1(pa[u],mx0[u]))upd1(pa[u],mx1[u]);
65         if(upd2(pa[u],mn0[u]))upd2(pa[u],mn1[u]);
66     }
67     for(int i=n-1;~i;i--){
68         ans1[i]+=ans1[i+1];
69         ans2[i]=max(ans2[i],ans2[i+1]);
70     }
71     for(int i=0;i<n;i++){
72         if(ans1[i])printf("%lld %lld\n",ans1[i],ans2[i]);
73         else puts("0 0");
74     }
75     return 0;
76 }

bzoj4199

【bzoj4199】【Noi2015】品酒大会

题解

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