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技术 2022年11月11日
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题意:给定一个长度为n的01串,你的任务是依次执行如表所示的m条指令:

1 p c 在第p个字符后插入字符,p = 0表示在整个字符串之前插入
2 p 删除第p个字符,后面的字符往前移
3 p1 p2反转第p1到第p2个字符
4 p1 p2输出从p1开始和p2开始的两个后缀的LCP。

析:对于前三个操作,splay 很容易就可以解决,但是对于最后一个操作,却不是那么容易,因为这是动态的,所以我们也要维护一个可以动态的,这就可以用Hash来解决,由于要翻转,所以要维护两个,一个正向的,一个反向的。在操作4时,先进行二分,然后用哈希进行判断,由于串不是太长,所以误差比较小。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e15;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 4e5 + 100;
const int mod = 3;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}// UVa 11996#define Key_value ch[ch[root][1]][0]
int pre[maxn], ch[maxn][2], key[maxn], sz[maxn];
int root, tot1;
int rev[maxn];
int s[maxn], tot2;
char a[maxn];
ULL H[maxn], revH[maxn];
ULL xp[maxn];void NewNode(int &rt, int fa, int x){
if(tot2) rt = s[tot2--];
else rt = ++tot1;
pre[rt] = fa;
key[rt] = H[rt] = revH[rt] = x;
ch[rt][0] = ch[rt][1] = 0;
rev[rt] = 0;
sz[rt] = 1;
}void push_up(int rt){
int l = ch[rt][0], r = ch[rt][1];
sz[rt] = sz[l] + sz[r] + 1;
H[rt] = key[rt] * xp[sz[r]] + H[r] + H[l] * xp[sz[r]+1];
revH[rt] = key[rt] * xp[sz[l]] + revH[l] + revH[r] * xp[sz[l]+1];
}void update_rev(int rt){
if(!rt) return ;
swap(ch[rt][0], ch[rt][1]);
swap(H[rt], revH[rt]);
rev[rt] ^= 1;
}void push_down(int rt){
if(rev[rt]){
update_rev(ch[rt][0]);
update_rev(ch[rt][1]);
rev[rt] = 0;
}
}void Build(int &rt, int l, int r, int fa){
if(l > r) return ;
int m = l+r >> 1;
NewNode(rt, fa, a[m] - '0');
Build(ch[rt][0], l, m-1, rt);
Build(ch[rt][1], m+1, r, rt);
push_up(rt);
}void Init(){
tot1 = root = tot2 = 0;
ch[root][0] = ch[root][1] = sz[root] = pre[root] = 0;
key[root] = 0; H[root] = revH[root] = 0;
NewNode(root, 0, -1);
NewNode(ch[root][1], root, -1);
scanf("%s", a);
Build(Key_value, 0, n-1, ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}int Get_kth(int rt, int k){
push_down(rt);
int t = sz[ch[rt][0]] + 1;
if(t == k) return rt;
if(t > k) return Get_kth(ch[rt][0], k);
return Get_kth(ch[rt][1], k-t);
}void Rotate(int x, int k){
int y = pre[x];
push_down(y);
push_down(x);
ch[y][!k] = ch[x][k];
pre[ch[x][k]] = y;
if(pre[y]) ch[pre[y]][ch[pre[y]][1]==y] = x;
pre[x] = pre[y];
ch[x][k] = y;
pre[y] = x;
push_up(y);
}void Splay(int rt, int goal){
push_down(rt);
while(pre[rt] != goal){
if(pre[pre[rt]] == goal){
push_down(pre[rt]);
push_down(rt);
Rotate(rt, ch[pre[rt]][0] == rt);
continue;
}
push_down(pre[pre[rt]]);
push_down(pre[rt]);
push_down(rt);
int y = pre[rt];
int k = ch[pre[y]][0] == y;
if(ch[y][k] == rt){
Rotate(rt, !k);
Rotate(rt, k);
}
else{
Rotate(y, k);
Rotate(rt, k);
}
}
push_up(rt);
if(goal == 0) root = rt;
}void Insert(int pos){
scanf("%s", a);
Splay(Get_kth(root, pos+1), 0);
Splay(Get_kth(root, pos+2), root);
Build(Key_value, 0, 0, ch[root][1]);
push_up(ch[root][1]);
push_up(root);
++n;
}void Erase(int rt){
if(!rt) return ;
s[++tot2] = rt;
Erase(ch[rt][0]);
Erase(ch[rt][1]);
}void Delete(int pos){
Splay(Get_kth(root, pos), 0);
Splay(Get_kth(root, pos+2), root);
Erase(Key_value);
pre[Key_value] = 0;
Key_value = 0;
push_up(ch[root][1]);
push_up(root);
--n;
}void Reverse(int pos, int tot){
Splay(Get_kth(root, pos), 0);
Splay(Get_kth(root, pos+tot+1), root);
update_rev(Key_value);
push_up(ch[root][1]);
push_up(root);
}bool judge(int p1, int p2, int mid){
Splay(Get_kth(root, p1), 0);
Splay(Get_kth(root, p1+mid+1), root);
ULL ans = H[Key_value];
Splay(Get_kth(root, p2), 0);
Splay(Get_kth(root, p2+mid+1), root);
return ans == H[Key_value];
}int solve(int p1, int p2){
int l = 1, r = n - p2 + 1;
while(l <= r){
int mid = l + r >> 1;
if(judge(p1, p2, mid)) l = mid + 1;
else r = mid - 1;
}
return l - 1;
}int main(){
xp[0] = 1;
for(int i = 1; i < maxn; ++i) xp[i] = xp[i-1] * mod;
while(scanf("%d %d", &n, &m) == 2){
Init();
while(m--){
int op, p, q;
scanf("%d %d", &op, &p);
if(1 == op) Insert(p);
else if(2 == op) Delete(p);
else if(3 == op){
scanf("%d", &q);
Reverse(p, q-p+1);
}
else{
scanf("%d", &q);
printf("%d\n", solve(p, q));
}
}
}
return 0;
}

  

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