BZOJ 1014 火星人 | 平衡树维护哈希

BZOJ 1014 火星人

题意

有一个字符串,三中操作:在某位置后面插入一个字符、修改某位置的字符、询问两个后缀的最长公共前缀。

题解

看到网上的dalao们都说这道题是平衡树,我就很懵x……平衡树维护什么啊?
最后发现某个节点维护的是它所代表的区间的哈希值——显然这个哈希值可以从左右子树的哈希值和这个节点上的字符算出来。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define space putchar(' ')
#define enter putchar('\n')
using namespace std;
typedef unsigned long long ll;
template <class T>
void read(T &x){
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
	if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
	x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x){
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
const int N = 100005, B = 29;
int n, m;
int idx, root, fa[N], ls[N], rs[N], val[N], sze[N];
ll hsh[N], pw[N];
char s[N];
#define which(x) (ls[fa[(x)]] == (x))
void upt(int u){
    sze[u] = sze[ls[u]] + sze[rs[u]] + 1;
    hsh[u] = hsh[ls[u]] + val[u] * pw[sze[ls[u]]] + hsh[rs[u]] * pw[sze[ls[u]] + 1];
}
void rotate(int u){
    int v = fa[u], w = fa[v], b = which(u) ? rs[u] : ls[u];
    if(w) which(v) ? ls[w] = u : rs[w] = u;
    which(u) ? (ls[v] = b, rs[u] = v) : (rs[v] = b, ls[u] = v);
    fa[u] = w, fa[v] = u;
    if(b) fa[b] = v;
    upt(v), upt(u);
}
void splay(int u, int tar){
    while(fa[u] != tar){
        if(fa[fa[u]] != tar){
            if(which(u) == which(fa[u])) rotate(fa[u]);
            else rotate(u);
        }
        rotate(u);
    }
    if(!tar) root = u;
}
int build(int l, int r, int pre){
    if(l > r) return 0;
    int u = ++idx, mid = (l + r) >> 1;
    val[u] = s[mid] - 'a' + 1, fa[u] = pre;
    ls[u] = build(l, mid - 1, u);
    rs[u] = build(mid + 1, r, u);
    upt(u);
    return u;
}
int find(int x){
    int u = root;
    while(sze[ls[u]] != x)
        if(x <= sze[ls[u]] - 1) u = ls[u];
        else x -= sze[ls[u]] + 1, u = rs[u];
    return u;
}
void insert(int pos, int x){
    int u = find(pos), v = find(pos + 1);
    splay(u, 0), splay(v, u);
    ls[v] = ++idx, fa[idx] = v, val[idx] = x, sze[idx] = 1;
    splay(idx, 0);
}
void change(int pos, int x){
    int u = find(pos);
    val[u] = x;
    splay(u, 0);
}
ll gethsh(int pos, int len){
    int u = find(pos - 1), v = find(pos + len);
    splay(u, 0), splay(v, u);
    return hsh[ls[v]];
}
int query(int a, int b){
    int l = 0, r = sze[root] - max(a, b) - 1, mid;
    while(l < r){
        mid = (l + r + 1) >> 1;
        if(gethsh(a, mid) == gethsh(b, mid)) l = mid;
        else r = mid - 1;
    }
    return l;
}
int main(){
    scanf("%s", s + 1);
    n = strlen(s + 1);
    pw[0] = 1;
    for(int i = 1; i < N; i++)
        pw[i] = pw[i - 1] * B;
    root = build(0, n + 1, 0);
    read(m);
    while(m--){
        char op[2];
        scanf("%s", op);
        if(op[0] == 'Q'){
            int a, b;
            read(a), read(b);
            write(query(a, b)), enter;
        }
        else if(op[0] == 'I'){
            int pos;
            scanf("%d%s", &pos, op);
            insert(pos, op[0] - 'a' + 1);
        }
        else if(op[0] == 'R'){
            int pos;
            scanf("%d%s", &pos, op);
            change(pos, op[0] - 'a' + 1);
        }
    }
    return 0;
}
本文作者:胡小兔
博客地址:http://rabbithu.cnblogs.com
posted @ 2017-12-08 13:24  胡小兔  阅读(299)  评论(0编辑  收藏  举报