hdu 3564 Another LIS （Splay Tree OR Segment Tree）

Problem - 3564

　　好不容易看见一道可以用Splay Tree写的题，结果应该是常数过大超时了。

　　好不容易还是给我找到bug了，就是一个字母打错，导致超时了。splay tree可以以421ms的时间完美通过。做法就是模拟插入数字，维护区间的最大LIS值即可。

Splay版代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int inf = 0x77777777;

struct Node {
    int data, cnt, max, val;
    Node *c[2], *p;
    Node(int data = 0, int val = 0, int cnt = 0) : data(data), val(val), cnt(cnt) {
        c[0] = c[1] = p = NULL;
        max = 0;
    }
} *Null, *Root, *Head, *Tail;

void up(Node *x) {
    Node *nl = x->c[0], *nr = x->c[1];
    x->max = max(nl->max, nr->max);
    x->max = max(x->max, x->val);
    x->cnt = nl->cnt + nr->cnt + 1;
}

void rotate(Node *x, bool right) {
    Node *y = x->p;
    bool left = !right;
    y->c[left] = x->c[right];
    if (x->c[right] != Null) x->c[right]->p = y;
    x->p = y->p;
    if (y->p != Null) {
        if (y->p->c[0] == y) {
            y->p->c[0] = x;
        } else {
            y->p->c[1] = x;
        }
    }
    x->c[right] = y;
    y->p = x;
    if (y == Root) Root = x;
    up(y);
}

void splay(Node *x, Node *f) {
    while (x->p != f) {
        if (x->p->p == f) {
            if (x->p->c[0] == x) {
                rotate(x, 1);
            } else {
                rotate(x, 0);
            }
        } else {
            Node *y = x->p, *z = y->p;
            if (z->c[0] == y) {
                if (y->c[0] == x) {
                    rotate(y, 1);
                    rotate(x, 1);
                } else {
                    rotate(x, 0);
                    rotate(x, 1);
                }
            } else {
                if (y->c[0] == x) {
                    rotate(x, 1);
                    rotate(x, 0);
                } else {
                    rotate(y, 0);
                    rotate(x, 0);
                }
            }
        }
    }
    up(x);
}

Node *getPos(int pos) {
//    cout << pos << "??" << endl;
    Node *p = Root;
    pos++;
//    cout << p->cnt << ' ' << p->data << ' ' << pos << endl;
//    cout << p->c[0]->cnt << ' ' << p->c[0]->data << ' ' << pos << endl;
    while (true) {
        int cnt = p->c[0]->cnt;
//        cout << cnt << ' ' << pos << endl;
        if (pos == cnt + 1) {
            break;
        } else if (pos <= cnt) {
            p = p->c[0];
        } else {
            pos -= cnt + 1;
            p = p->c[1];
        }
    }
    return p;
}

void inTra(Node *x) {
    if (x == Null) return ;
    inTra(x->c[0]);
    cout << x->data << ' ';
    inTra(x->c[1]);
}

void insPos(int elem, int val, int pos) {
//    cout << "here " << pos << endl;
//    inTra(Root);
    Node *pl = getPos(pos), *pr = getPos(pos + 1);
//    cout << pl->data << "= =" << pr->data << " ~~ " << pl->cnt << "!!" << pr->cnt << endl;
    Node *tmp = new Node(elem, val, 1);
    splay(pl, Null);
    splay(pr, pl);
    pr->c[0] = tmp;
    tmp->p = pr;
    tmp->c[0] = tmp->c[1] = Null;
    splay(tmp, Null);
}

void init() {
    Null = new Node();
    Head = Root = new Node(inf, 0, 2);
    Tail = new Node(inf, 0, 1);
    Root->c[1] = Tail;
    Root->c[0] = Root->p = Null;
    Tail->p = Root;
    Tail->c[0] = Tail->c[1] = Null;
}

int getMax(int pos) {
    Node *pl = getPos(0), *pr = getPos(pos + 1);
//    cout << pl->data << "= =" << pr->data << endl;
    splay(pl, Null);
    splay(pr, pl);
//    inTra(Root);
//    cout << "tra~~" << endl;
    pr = pr->c[0];
    if (pr == Head || pr == Tail) return 0;
    return pr->max;
}

int main() {
//    freopen("in", "r", stdin);
    int n, T, x;
    scanf("%d", &T);
    for (int cas = 1; cas <= T; cas++) {
        init();
        scanf("%d", &n);
        printf("Case #%d:\n", cas);
        int LIS = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &x);
            int tmp = getMax(x);
//            cout << "tmp " << tmp << endl;
            LIS = max(LIS, tmp + 1);
            printf("%d\n", LIS);
//            cout << "can not insert?" << endl;
            insPos(i, tmp + 1, x);
//            cout << "no!!" << endl;
//            inTra(Root);
//            puts("");
        }
        puts("");
    }
    return 0;
}

线段树的之后补上。

UPD:

　　解释一下我的线段树法的操作。首先，我把插入的位置倒序插入线段树中，从而得到每个数最终的位置。然后再正序模拟插入过程，更新LIS的值。

线段树版本：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

const int N = 111111;
int mx[N << 2], cnt[N << 2];

void up1(int rt) {
    cnt[rt] = cnt[rt << 1] + cnt[rt << 1 | 1];
}

void build1(int l, int r, int rt) {
    if (l == r) {
        cnt[rt] = 1;
        return ;
    }
    int m = l + r >> 1;
    build1(lson);
    build1(rson);
    up1(rt);
}

int ins1(int x, int l, int r, int rt) {
    if (l == r) {
        cnt[rt] = 0;
        return l;
    }
    int m = l + r >> 1, ret;
    if (cnt[rt << 1] >= x) {
        ret = ins1(x, lson);
    } else {
        ret = ins1(x - cnt[rt << 1], rson);
    }
    up1(rt);
    return ret;
}

int rec[N], id[N];

void PRE(int n) {
    for (int i = 0; i < n; i++) scanf("%d", &rec[i]);
    build1(0, n - 1, 1);
    for (int i = n - 1; i >= 0; i--) id[i] = ins1(rec[i] + 1, 0, n - 1, 1);
//    for (int i = 0; i < n; i++) cout << i << " : " << id[i] << ' ' << rec[i] << endl;
}

void up2(int rt) {
    mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}

void build2(int l, int r, int rt) {
    cnt[rt] = mx[rt] = 0;
    if (l == r) return ;
    int m = l + r >> 1;
    build2(lson);
    build2(rson);
}

void ins2(int x, int val, int l, int r, int rt) {
    if (l == r) {
        mx[rt] = val;
        return ;
    }
    int m = l + r >> 1;
    if (x <= m) ins2(x, val, lson);
    else ins2(x, val, rson);
    up2(rt);
}

int query2(int L, int R, int l, int r, int rt) {
    if (L <= l && r <= R) {
        return mx[rt];
    }
    int m = l + r >> 1, ret = 0;
    if (L <= m) ret = max(ret, query2(L, R, lson));
    if (m < R) ret = max(ret, query2(L, R, rson));
    return ret;
}

void work(int n) {
    int tmp, curMax = 0;
    build2(0, n - 1, 1);
    for (int i = 0; i < n; i++) {
        tmp = query2(0, id[i], 0, n - 1, 1);
        curMax = max(curMax, tmp + 1);
        printf("%d\n", curMax);
        ins2(id[i], tmp + 1, 0, n - 1, 1);
    }
}

int main() {
//    freopen("in", "r", stdin);
    int T, n;
    scanf("%d", &T);
    for (int i = 1; i <= T; i++) {
        scanf("%d", &n);
        printf("Case #%d:\n", i);
        PRE(n);
        work(n);
        puts("");
    }
    return 0;
}

——written by Lyon