bzoj1396: 识别子串

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #define maxn 100005
 #define inf 100000000
 using namespace std;
 char st[maxn];
 int n,tot,last,root,sum[maxn<<],tmp[maxn<<],fa[maxn<<],son[maxn<<][],dist[maxn<<],ri[maxn<<],pos[maxn<<];
 struct Tsegment{
     void prepare(){tot=last=root=,memset(ri,,sizeof(ri));}
     int newnode(int x){
         dist[++tot]=x; return tot;
     }
     void add(int op,int x){
         int p=last,np=newnode(dist[p]+); last=np; ri[np]=; pos[np]=op;
         for (;p&&!son[p][x];p=fa[p]) son[p][x]=np;
         if (p==) fa[np]=root;
         else{
             int q=son[p][x];
             if (dist[p]+==dist[q]) fa[np]=q;
             else{
                 int nq=newnode(dist[p]+);
                 memcpy(son[nq],son[q],sizeof(son[q]));
                 fa[nq]=fa[q],fa[q]=fa[np]=nq;
                 for (;p&&son[p][x]==q;p=fa[p]) son[p][x]=nq;
             }
         }
     }
 }SAM;
 struct Fsegment{
     int l,r,lazy,val;
 }tree1[maxn*];
 struct date{
     void build(int k,int l,int r){
         tree1[k].lazy=tree1[k].val=inf; tree1[k].l=l,tree1[k].r=r;
         if (l==r) return;
         int mid=(l+r)/;
         build(k*,l,mid),build(k*+,mid+,r);
     }
     void change(int k,int l,int r,int x,int y,int z){
         if (x>y) return;
         if (tree1[k].lazy!=inf){
             if (tree1[k*].l) pushdown(k*,tree1[k].lazy);
             if (tree1[k*+].l) pushdown(k*+,tree1[k].lazy);
             tree1[k].lazy=inf;
         }
         if (l>=x&&r<=y){
             pushdown(k,z);
             return;
         } int mid=(l+r)/;
         if (x<=mid) change(k*,l,mid,x,y,z);
         if (y>mid) change(k*+,mid+,r,x,y,z);
     }
     void pushdown(int k,int x){
         tree1[k].lazy=min(tree1[k].lazy,x);
         if (tree1[k].l==tree1[k].r) tree1[k].val=min(tree1[k].val,tree1[k].lazy);
     }
     int query(int k,int l,int r,int x){
         if (tree1[k].lazy!=inf){
             if (tree1[k*].l) pushdown(k*,tree1[k].lazy);
             if (tree1[k*+].l) pushdown(k*+,tree1[k].lazy);
             tree1[k].lazy=inf;
         }
         if (l==r&&r==x) return tree1[k].val;
         int mid=(l+r)>>,ans=inf;
         if (x<=mid) ans=min(ans,query(k*,l,mid,x));
         else ans=min(ans,query(k*+,mid+,r,x));
         return ans;
     }
 }Tree1;
 struct Ksegment{
     int l,r,val,lazy;
 }tree[maxn*];
 struct Graph{
     void build(int k,int l,int r){
         tree[k].l=l,tree[k].r=r,tree[k].lazy=tree[k].val=inf;
         if (l==r) return; int mid=(l+r)/;
         build(k*,l,mid),build(k*+,mid+,r);
     }
     void change(int k,int l,int r,int x,int y,int z){
         if (x>y) return;
         if (tree[k].lazy!=inf){
             if (tree[k*].l) pushdown(k*,tree[k].lazy);
             if (tree[k*+].l) pushdown(k*+,tree[k].lazy);
             tree[k].lazy=inf;
         }
         if (l>=x&&r<=y){
             pushdown(k,z);
             return;
         } int mid=(l+r)>>;
         if (x<=mid) change(k*,l,mid,x,y,z);
         if (y>mid) change(k*+,mid+,r,x,y,z);
     }
     void pushdown(int k,int x){
         tree[k].lazy=min(tree[k].lazy,x);
         if (tree[k].l==tree[k].r) tree[k].val=min(tree[k].val,tree[k].lazy);
     }
     int query(int k,int l,int r,int x){
         if (tree[k].lazy!=inf){
             if (tree[k*].l) pushdown(k*,tree[k].lazy);
             if (tree[k*+].r) pushdown(k*+,tree[k].lazy);
             tree[k].lazy=inf;
         }
         if (l==r&&r==x) return tree[k].val;
         int mid=(l+r)>>,ans=inf;
         if (x<=mid) ans=min(ans,query(k*,l,mid,x));
         else ans=min(ans,query(k*+,mid+,r,x));
         return ans;
     }
 }Tree;
 int main(){
     scanf("%s",st+),n=strlen(st+);
     SAM.prepare();
     for (int i=;i<=n;i++) SAM.add(i,st[i]-'a');
     memset(sum,,sizeof(sum));
     for (int i=;i<=tot;i++) sum[dist[i]]++;
     for (int i=;i<=tot;i++) sum[i]+=sum[i-];
     for (int i=;i<=tot;i++) tmp[sum[dist[i]]--]=i;
     for (int i=tot,x;i>=;i--){
         x=tmp[i];
         if (fa[x]) ri[fa[x]]+=ri[x];
     }
     ri[root]=;
     Tree1.build(,,n);
     Tree.build(,,n);
     for (int i=;i<=tot;i++){
         if (ri[i]!=) continue;
         int x=pos[i],y=dist[i],z=dist[fa[i]]+;
         Tree1.change(,,n,x-y+,x-z+,x);   //第一棵线段树按位置,记得减
         Tree.change(,,n,x-z+,x,z);    //第二棵线段树按长度
     }
     for (int i=;i<=n;i++){
         printf("%d\n",min(Tree1.query(,,n,i)-i+,Tree.query(,,n,i)));
     }
     return ;
 }
 

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1396

题目大意：

做法；看到题目中所说的T在S中只出现过一次，就很容易想到用后缀自动机嘛，显然就是right值为一的状态，而且right值为一的状态只能是每次add时第一个新建的点，这很显然嘛，这就很方便记录了。然后再用线段树维护一下最小值，稍微想一下就行，当时我竟然是很快就想到了，不过我inf开小了，狂WA不止。

后缀自动机+线段树。