洛谷 - P4008 - 文本编辑器 - 无旋Treap

https://www.luogu.org/problem/P4008

无旋Treap也可以维护序列。
千万要注意要先判断p节点存在才进行Show操作,不然输出一个'\0'(或者RecBin里面的东西)草。

假如有限制同时存在的节点数量的话,UnBuild操作是显得重要的。

当然这里最主要的是类似笛卡尔树的O(n)建立Treap树。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ls(p) ch[p][0]
#define rs(p) ch[p][1]

const int MAXN = 2000000 + 5;
char val[MAXN];
int ch[MAXN][2], rnd[MAXN], siz[MAXN], tot, root;

int cur;
stack<int> RecBin;

inline void Init() {
    root = 0, tot = 0;
    cur = 0;
    srand(19260817);
    while(RecBin.size())
        RecBin.pop();
}

inline void PushUp(int p) {
    siz[p] = siz[ls(p)] + siz[rs(p)] + 1;
}

void SplitRank(int p, int rk, int &x, int &y) {
    if(!p) {
        x = y = 0;
        return;
    }
    //PushDown(p);
    if(rk <= siz[ls(p)]) {
        y = p;
        SplitRank(ls(p), rk, x, ls(p));
        PushUp(y);
    } else {
        x = p;
        SplitRank(rs(p), rk - siz[ls(p)] - 1, rs(p), y);
        PushUp(x);
    }
}

int Merge(int x, int y) {
    if(!x || !y)
        return x | y;
    //这个是小根Treap
    if(rnd[x] < rnd[y]) {
        //PushDown(x);
        rs(x) = Merge(rs(x), y);
        PushUp(x);
        return x;
    } else {
        //PushDown(y);
        ls(y) = Merge(x, ls(y));
        PushUp(y);
        return y;
    }
}

inline int NewNode(char v) {
    int p;
    if(RecBin.size()) {
        p = RecBin.top();
        RecBin.pop();
    } else
        p = ++tot;
    ch[p][0] = ch[p][1] = 0;
    val[p] = v;
    rnd[p] = rand();
    siz[p] = 1;
    return p;
}

void Show(int p) {
    if(!p)
        return;
    Show(ls(p));
    putchar(val[p]);
    Show(rs(p));
}

//O(n)建树,返回新树的根
int st[MAXN], stop;
char buf[MAXN];
inline int Build(int n) {
    stop = 0;
    for(int i = 0; i < n; ++i) {
        int tmp = NewNode(buf[i]), last = 0;
        while(stop && rnd[st[stop]] > rnd[tmp]) {
            last = st[stop];
            PushUp(last);
            st[stop--] = 0;
        }
        if(stop)
            rs(st[stop]) = tmp;
        ls(tmp) = last;
        st[++stop] = tmp;
    }
    while(stop)
        PushUp(st[stop--]);
    return st[1];
}

//O(n)回收整棵树
inline void UnBuild(int p) {
    if(!p)
        return;
    UnBuild(ls(p));
    UnBuild(rs(p));
    RecBin.push(p);
}

inline void Move() {
    scanf("%d", &cur);
}

inline void Insert(int &root) {
    int x = 0, y = 0, z = 0, n;
    SplitRank(root, cur, x, z);
    scanf("%d", &n);
    buf[n] = '\0';
    for(int i = 0; i < n; ++i) {
        char ch = getchar();
        while(ch < 32 || ch > 126 || ch == '\r' || ch == '\n')
            ch = getchar();
        buf[i] = ch;
    }
    y = Build(n);
    root = Merge(Merge(x, y), z);
}

inline void Delete(int &root) {
    int x = 0, y = 0, z = 0, n;
    SplitRank(root, cur, x, y);
    scanf("%d", &n);
    SplitRank(y, n, y, z);
    //会不会太慢了
    UnBuild(y);
    //y=Merge(ls(y),rs(y));
    root = Merge(x, z);
}

inline void Get(int &root) {
    int x = 0, y = 0, z = 0, n = 0;
    SplitRank(root, cur, x, y);
    scanf("%d", &n);
    SplitRank(y, n, y, z);
    Show(y);
    puts("");
    root = Merge(Merge(x, y), z);
}

inline void Prev() {
    --cur;
}

inline void Next() {
    ++cur;
}

char op[20];
int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int t;
    scanf("%d", &t);
    Init();
    while(t--) {
        scanf("%s ", op);
        switch(op[0]) {
            case 'M':
                Move();
                break;
            case 'I':
                Insert(root);
                break;
            case 'D':
                Delete(root);
                break;
            case 'G':
                Get(root);
                break;
            case 'P':
                Prev();
                break;
            case 'N':
                Next();
                break;
        }
    }
    return 0;
}
posted @ 2019-08-10 01:33  韵意  阅读(211)  评论(0编辑  收藏  举报