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POJ 3468.A Simple Problem with Integers 解题报告

用树状数组和线段树会比较简单,这里用这道题来学习Splay。

第一次写,代码比较丑

/*       初始化添加一个key值足够大的结点       保证每个需要的结点都有后继*/#include <iostream>#include <cstdio>#define ll long longusing namespace std;const int MAXN = 1111111, INF = 0x7fffffff;struct node {    //需要的记录信息    ll  key, val, sum, lazy, Size, Cnt;    //指向儿子和父亲的指针    node *ch[2], *pre;    node() {pre = ch[0] = ch [1] = 0; Size = 1; key = 0;}    node (ll key) : key (key) {pre = ch[0] = ch[1] = 0; Size = 1, Cnt = 1, lazy = 0;}    void Csize() {        Size = Cnt;        if (ch[0] != NULL) Size += ch[0]->Size;        if (ch[1] != NULL) Size += ch[1]->Size;    }    void Csum() {        sum = val;        if (ch[0] != NULL) sum += ch[0]->sum + ch[0]->Size * ch[0]->lazy;        if (ch[1] != NULL) sum += ch[1]->sum + ch[1]->Size * ch[1]->lazy;    }} nil (0), *NIL = &nil;struct Splay {    node *root, nod[MAXN];    int ncnt;//计算key值不同的结点数,已去重    Splay() {        ncnt = 0;        root = & (nod[ncnt++] = node (INF) );        root->pre = NIL;        root->val = root->sum = 0;    }    void Push_Down (node *x) {        if (x->lazy != 0) {            if (x->ch[0] != NULL) x->ch[0]->lazy += x->lazy;            if (x->ch[1] != NULL) x->ch[1]->lazy += x->lazy;                     x->val+=x->lazy;        }        x->lazy = 0;    }    void Update (node *x) {        x->Csize();        x->Csum();    }    void Rotate (node *x, int sta) { //单旋转操作,0左旋,1右旋        node *p = x->pre, *g = p->pre;        Push_Down (p), Push_Down (x);        p->ch[!sta] = x->ch[sta];        if (x->ch[sta] != NULL)  x->ch[sta]->pre = p;        x->pre = g;        if (g != NIL)            if (g->ch[0] == p)  g->ch[0] = x;            else g->ch[1] = x;        x->ch[sta] = p, p->pre = x, Update (p);        if (p == root ) root = x;    }    void splay (node *x, node *y) { //Splay 操作,表示把结点x,转到根        for (Push_Down (x) ; x->pre != y;) { //将x的标记往下传            if (x->pre->pre == y) { //目标结点为父结点                if (x->pre->ch[0] == x)       Rotate (x, 1);                else   Rotate (x, 0);            }            else {                node *p = x->pre, *g = p->pre;                if (g->ch[0] == p)                    if (p->ch[0] == x)                        Rotate (p, 1), Rotate (x, 1);//   / 一字型双旋转                    else                        Rotate (x, 0), Rotate (x, 1);//  < 之字形双旋转                else if (p ->ch[1] == x)                    Rotate (p, 0), Rotate (x, 0);//    \ 一字型旋转                else                    Rotate (x, 1), Rotate (x, 0); //   >之字形旋转            }        }        Update (x); //维护x结点    }    //找到中序便利的第K个结点,并旋转至结点y的下面。    void Select (int k, node *y) {        int tem ;        node *t ;        for ( t = root; ; ) {            Push_Down (t) ; //标记下传            tem = t->ch[0]->Size ;            if (k == tem + 1 ) break ; //找到了第k个结点 t            if (k <= tem)                t = t->ch[0] ; //第k个结点在左子树            else                k -= tem + 1 , t = t->ch[1] ;//在右子树        }        splay (t, y);    }    bool Search (ll key, node *y) {        node *t = root;        for (; t != NULL;) {            Push_Down (t);            if (t->key > key && t->ch[0] != NULL) t = t->ch[0];            else if (t->key < key && t->ch[1] != NULL) t = t->ch[1];            else                break;        }        splay (t, y);        return t->key == key;    }    void Insert (int key, int val) {        if (Search (key, NIL) ) root->Cnt++, root->Size++;        else {            int d = key > root->key;            node *t = & (nod[++ncnt] = node (key) );            Push_Down (root);            t->val = t->sum = val;            t->ch[d] = root->ch[d];            if (root->ch[d] != NULL) root->ch[d]->pre = t;            t->ch[!d] = root;            t->pre = root->pre;            root->pre = t;            root->ch[d] = NULL;            Update (root);            root = t;        }        Update (root);    }} sp;ll  n, m, x;int main() {    scanf ("%lld %lld", &n, &m);    sp.Insert (0, 0);    sp.Insert (n + 1, 0);    for (int i = 1; i <= n; i++) {        scanf ("%lld", &x);        sp.Insert (i, x);    }    char cmd;    int l, r;    for (int i = 1; i <= m; i++) {        scanf ("\n%c %d %d", &cmd, &l, &r);        sp.Search (l - 1, NIL);        sp.Search (r + 1, sp.root);        if (cmd == Q) {            node *t = sp.root->ch[1]->ch[0];            ll ans = t->sum + t->Size * t->lazy;            printf ("%lld\n", ans);        }        if (cmd == C) {            ll c;            scanf ("%lld", &c);            sp.root->ch[1]->ch[0]->lazy += c;        }    }    return 0;}
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POJ 3468.A Simple Problem with Integers 解题报告