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  • UVa 11996 Jewel Magic (splay + Hash + 二分)

    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
    ", solve(p, q));
          }
        }
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7455265.html
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