2555: SubString

2555: SubString

Time Limit: 30 Sec  Memory Limit: 512 MB
Submit: 1910  Solved: 539
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Description

  
    懒得写背景了，给你一个字符串init，要求你支持两个操作
    
    (1):在当前字符串的后面插入一个字符串
    
    (2):询问字符串s在当前字符串中出现了几次？(作为连续子串)
    
    你必须在线支持这些操作。
    

Input

    第一行一个数Q表示操作个数
    
    第二行一个字符串表示初始字符串init
    
    接下来Q行，每行2个字符串Type,Str 
    
    Type是ADD的话表示在后面插入字符串。
    
    Type是QUERY的话表示询问某字符串在当前字符串中出现了几次。
    
    为了体现在线操作，你需要维护一个变量mask，初始值为0
   
    
    读入串Str之后，使用这个过程将之解码成真正询问的串TrueStr。
    询问的时候，对TrueStr询问后输出一行答案Result
    然后mask = mask xor Result  
    插入的时候，将TrueStr插到当前字符串后面即可。

HINT:ADD和QUERY操作的字符串都需要解压
   

Output

Sample Input

2

A

QUERY B

ADD BBABBBBAAB

Sample Output

0

HINT

 40 % 的数据字符串最终长度 <= 20000，询问次数<= 1000，询问总长度<= 10000

    

100 % 的数据字符串最终长度 <= 600000，询问次数<= 10000,询问总长度<= 3000000

新加数据一组--2015.05.20

   

Source

Ctsc模拟赛By 洁妹

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使用后缀自动机+LCT解决，，传送门后缀自动机方面看CLJ的ppt啦。。。后缀自动机里每个节点存一个子串，该串是从init开始，每次走添加字符边形成的，新字符放在后面这个点同时维护这个子串在母串中出现位置的集合然后定义parent边，par(s)为字符串x，满足s出现集合是x集合的真子集，且x的size最小每个点都有一个长度，对于节点s，该串的长度在(len(par(s)),len(s)]的后缀都满足当前点维护的位置集合构造方法见论文吧。。。。。因为这个后缀自动机是个动态结构，所以可以用LCT维护

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E6 + 2E5 + 233;int n,Q,mas,rt,last,cnt,len[maxn],par[maxn],g[maxn][26],sum[maxn],fa[maxn],pfa[maxn],ch[maxn][2],rev[maxn],Add[maxn];char c[maxn];stack <int> st;void pushdown(int x){if (rev[x]) {swap(ch[x][0],ch[x][1]);rev[x] ^= 1;for (int i = 0; i < 2; i++)if (ch[x][i]) rev[ch[x][i]] ^= 1;}if (Add[x] != 0){sum[x] += Add[x];for (int i = 0; i < 2; i++)if (ch[x][i] != 0) Add[ch[x][i]] += Add[x];Add[x] = 0;}}void rotate(int x){int y = fa[x],z = fa[y];int d = (ch[y][0] == x)?0:1;fa[x] = z; fa[y] = x;pfa[x] = pfa[y]; pfa[y] = 0;ch[y][d] = ch[x][d^1];if (ch[y][d]) fa[ch[y][d]] = y;ch[x][d^1] = y;if (z) ch[z][ch[z][1] == y] = x;}void splay(int x){for (int z = x; z; z = fa[z]) st.push(z);while (!st.empty()) pushdown(st.top()),st.pop();for (int y = fa[x]; y; rotate(x),y = fa[x])if (fa[y])rotate((ch[fa[y]][0] == y)^(ch[y][0] == x)?x:y);}void Access(int x){for (int v = 0; x; v = x,x = pfa[x]){splay(x);if (ch[x][1]) fa[ch[x][1]] = 0,pfa[ch[x][1]] = x;if (v != 0) pfa[v] = 0,fa[v] = x;ch[x][1] = v;}}void ChangeRoot(int x){Access(x); splay(x);rev[x] ^= 1;}void Join(int x,int y){ChangeRoot(x); pfa[x] = y;Access(x); splay(x);}void Cut(int x,int y){ChangeRoot(x);Access(y); splay(y);fa[x] = ch[y][0] = 0;}void LCT_join(int x,int y){Join(x,y); ChangeRoot(rt);Access(y); splay(y);sum[y] += sum[x];if (ch[y][0]) Add[ch[y][0]] += sum[x]; }void LCT_cut(int x,int y){Cut(x,y); ChangeRoot(rt);Access(y); splay(y);sum[y] -= sum[x];if (ch[y][0]) Add[ch[y][0]] -= sum[x];}int New_node(int le,int va){int ret = ++cnt; len[ret] = le;sum[ret] = va; return ret;}void Extend(int w){int p = last,np = New_node(len[p] + 1,1);while (p && !g[p][w])g[p][w] = np,p = par[p];if (!p){par[np] = rt;LCT_join(np,rt);last = np; return;}int q = g[p][w];if (len[p] + 1 == len[q])par[np] = q,LCT_join(np,q);else{int nq = New_node(len[p] + 1,0);for (int i = 0; i < 26; i++) g[nq][i] = g[q][i];par[nq] = par[q]; par[q] = par[np] = nq;LCT_cut(q,par[nq]); LCT_join(q,nq);LCT_join(np,nq); LCT_join(nq,par[nq]);while (p && g[p][w] == q)g[p][w] = nq,p = par[p];}last = np;}int Getcom(int mas){scanf("%s",c);int ret = (c[0] == 'A')?1:2;scanf("%s",c); n = strlen(c);for (int i = 0; i < n; i++){mas = (mas*131 + i) % n;swap(c[i],c[mas]);}return ret;}int main(){last = rt = New_node(0,0); cin >> Q; scanf("%s",c); n = strlen(c);for (int i = 0; i < n; i++) Extend(c[i] - 'A');while (Q--){int typ = Getcom(mas);if (typ == 1){for (int i = 0; i < n; i++) Extend(c[i] - 'A');}else{int s = rt;for (int i = 0; i < n && s; i++) s = g[s][c[i] - 'A'];if (!s) {puts("0"); continue;}Access(s); splay(s);mas ^= sum[s];printf("%d\n",sum[s]);}}return 0;}