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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;}
POJ 3468.A Simple Problem with Integers 解题报告
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