• poj 3580 SuperMemo


    题目连接

    http://poj.org/problem?id=3580  

    SuperMemo

    Description

    Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1A2, ... An}. Then the host performs a series of operations and queries on the sequence which consists:

    1. ADD x y D: Add D to each number in sub-sequence {Ax ... Ay}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
    2. REVERSE x y: reverse the sub-sequence {Ax ... Ay}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
    3. REVOLVE x y T: rotate sub-sequence {Ax ... AyT times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
    4. INSERT x P: insert P after Ax. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
    5. DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
    6. MIN x y: query the participant what is the minimum number in sub-sequence {Ax ... Ay}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2

    To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.

    Input

    The first line contains (≤ 100000).

    The following n lines describe the sequence.

    Then follows M (≤ 100000), the numbers of operations and queries.

    The following M lines describe the operations and queries.

    Output

    For each "MIN" query, output the correct answer.

    Sample Input

    6
    1
    7
    8
    5
    2
    9
    8
    ADD 1 3 2
    INSERT 4 7
    REVERSE 1 7
    MIN 3 5
    DEL 2
    DEL 3
    REVOLVE 1 5 3
    MIN 1 3

    Sample Output

    5
    3

    伸展树模板题,几乎所有的操作都在里面了。。

    #include<algorithm>
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<set>
    using std::set;
    using std::min;
    using std::pair;
    using std::swap;
    using std::vector;
    using std::multiset;
    #define pb(e) push_back(e)
    #define sz(c) (int)(c).size()
    #define mp(a, b) make_pair(a, b)
    #define all(c) (c).begin(), (c).end()
    #define iter(c) __typeof((c).begin())
    #define cls(arr, val) memset(arr, val, sizeof(arr))
    #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    #define rep(i, n) for(int i = 0; i < (int)n; i++)
    #define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
    const int N = 1 << 17;
    const int INF = ~0u >> 1;
    int arr[N];
    struct Node {
        bool rev;
        int dat, s, mval, delta;
        Node *fa, *ch[2];
        inline void set(int _dat, int _s, Node *p) {
            rev = delta =0;
            dat = mval = _dat, s = _s;
            fa = ch[0] = ch[1] = p;
        }
        inline bool dir() const {
            return this == fa->ch[0];
        }
        inline void link(Node *x, bool d) {
            ch[d] = x, x->fa = this;
        }
        inline void push_up() {
            s = ch[0]->s + ch[1]->s + 1;
            mval = min(dat, min(ch[0]->mval, ch[1]->mval));
        }
        inline void update() {
            if(!s) return;
            rev ^= 1;
            swap(ch[0], ch[1]);
        }
        inline void add(int v) {
            if(!s) return;
            delta += v;
            mval += v;
            dat += v;
        }
        inline void push_down() {
            if(delta) {
                ch[0]->add(delta);
                ch[1]->add(delta);
                delta = 0;
            }
            if(rev) {
                ch[0]->update();
                ch[1]->update();
                rev ^= 1;
            }
        }
    };
    #define RRL root->ch[1]->ch[0]
    struct SplayTree {
        int top;
        Node *root, *tail;
        Node *null, stack[N], *pool[N >> 1];
        inline void init(int n) {
            top = 0, tail = &stack[0];
            null = tail++;
            null->set(INF, 0, NULL);
            root = newNode(INF);
            root->link(newNode(INF), 1);
            Node *x = built(1, n);
            root->ch[1]->link(x, 0);
            root ->ch[1]->push_up();
            root->push_up();
            splay(x, null);
        }
        inline Node *newNode(int v) {
            Node *p = !top ? tail++ : pool[--top];
            p->set(v, 1, null);
            return p;
        }
        inline Node *built(int l, int r) {
            if(l > r) return null;
            int mid = (l + r) >> 1;
            Node *p = newNode(arr[mid]);
            p->ch[0] = built(l, mid - 1);
            if(p->ch[0]->s) p->ch[0]->fa = p;
            p->ch[1] = built(mid + 1, r);
            if(p->ch[1]->s) p->ch[1]->fa = p;
            p->push_up();
            return p;
        }
        inline void rotate(Node *&x, bool d) {
            Node *y = x->fa;
            y->push_down(), x->push_down();
            y->ch[!d] = x->ch[d];
            if(x->ch[d]->s) x->ch[d]->fa = y;
            x->fa = y->fa;
            if(y->fa->s) y->fa->ch[!y->dir()] = x;
            x->ch[d] = y;
            y->fa = x, y->push_up();
            if(y == root) root = x;
        }
        inline void splay(Node *x, Node *f) {
            for(;x->fa !=f ;x->push_down()) {
                if(x->fa->fa == f) {
                    rotate(x, x->dir());
                } else {
                    bool d = x->fa->dir();
                    if(d == x->dir()) rotate(x->fa, d), rotate(x, d);
                    else rotate(x, !d), rotate(x, d);
                }
            }
            x->push_up();
        }
        inline Node *select(Node *x, int k) {
            for(int t = 0; x->s; ) {
                x->push_down();
                t = x->ch[0]->s;
                if(t == k) break;
                if(k < t) x = x->ch[0];
                else k -= t + 1, x = x->ch[1];
            }
            return x;
        }
        inline void get_range(int l, int r) {
            splay(select(root, l - 1), null);
            splay(select(root, r + 1), root);
        }
        inline void add(int l, int r, int v) {
            get_range(l, r);
            RRL->add(v);
            splay(RRL, null);
        }
        inline void insert(int k, int v) {
            splay(select(root, k), null);
            splay(select(root, k + 1), root);
            root->ch[1]->link(newNode(v), 0);
            root->ch[1]->push_up();
            root->push_up();
            splay(RRL, null);
        }
        inline void reverse(int l, int r) {
            get_range(l, r);
            RRL->update();
        }
        inline void revolve(int l, int r, int k) {
            int len = (r - l + 1);
            k = (k % len + len) % len;
            if(!k) return;
            if(1 == k) {
                erase(r);
                insert(l - 1, pool[--top]->dat);
            } else {
                splay(select(root, r - k + 1), null);
                splay(select(root, r + 1), root);
                splay(select(root, l - 1), root);
                splay(select(root, r), root->ch[1]);
                Node *x = root->ch[0]->ch[1];
                root->ch[0]->ch[1] = null;
                root->ch[0]->push_up();
                root->ch[1]->ch[0]->link(x, 1);
                splay(x, null);
            }
        }
        inline void erase(int k) {
            splay(select(root, k), null);
            Node *ret = root;
            root = root->ch[1];
            root->fa = null;
            splay(select(root, 0), null);
            root->ch[0] = ret->ch[0];
            root->ch[0]->fa = root;
            root->push_up();
            pool[top++] = ret;
        }
        inline void query(int l, int r) {
            get_range(l, r);
            printf("%d
    ", RRL->mval);
        }
    }spt;
    int main() {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w+", stdout);
    #endif
        char buf[20];
        int n, q, a, b, c;
        while(~scanf("%d", &n)) {
            rep(i, n) scanf("%d", &arr[i + 1]);
            spt.init(n);
            scanf("%d", &q);
            while(q--) {
                scanf("%s", buf);
                if('D' == buf[2]) {
                    scanf("%d %d %d", &a, &b, &c);
                    spt.add(a, b, c);
                } else if('S' == buf[2]) {
                    scanf("%d %d", &a, &b);
                    spt.insert(a, b);
                } else if('V' == buf[2] && 'E' == buf[3]) {
                    scanf("%d %d", &a, &b);
                    spt.reverse(a, b);
                } else if('V' == buf[2] && 'O' == buf[3]) {
                    scanf("%d %d %d", &a, &b, &c);
                    spt.revolve(a, b, c);
                } else if('L' == buf[2]) {
                    scanf("%d", &a);
                    spt.erase(a);
                } else if('N' == buf[2]) {
                    scanf("%d %d", &a, &b);
                    spt.query(a, b);
                }
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/GadyPu/p/4773536.html
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