poj 3580 SuperMemo

题目传送门

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, {A₁, A₂, ... A_n}. 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 {A_x ... A_y}. 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 {A_x ... A_y}. 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 {A_x ... A_y} T 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 A_x. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
5. DELETE x: delete A_x. 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 {A_x ... A_y}. 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 n (n ≤ 100000).

The following n lines describe the sequence.

Then follows M (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

51 2 3 4 52ADD 2 4 1MIN 4 5

Sample Output

5

题解

这道题是典型的多操作题，调起来特别费时间。这道题我前天开始做，今天才打完，报了好多次RE。题目其实挺简单的就是难调。我在序列中加入了两个哨兵（赋极大值），避免对整棵树进行操作，无法获得区间。

#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>using namespace std;inline int getint() {int a = 0; char c = getchar();while (c < '0' || c > '9') c = getchar();while (c >= '0' && c <= '9') {a = (a << 3) + (a << 1) + c - '0';c = getchar();}return a;}struct Node {int v, ly, m, s;bool rot;Node *fa, *ch[2];Node() : v(0), fa(NULL), ly(0), rot(false), m(0), s(0) { memset(ch, 0, sizeof ch); }void pushdown();void maintain();}NN[200005], *root;int totNN = -1, a[100005];void Node::pushdown() {if (rot) {if (ch[0]) ch[0]->rot ^= 1, swap(ch[0]->ch[0], ch[0]->ch[1]);if (ch[1]) ch[1]->rot ^= 1, swap(ch[1]->ch[0], ch[1]->ch[1]);rot = false;}if (ly) {if (ch[0]) ch[0]->v += ly, ch[0]->m += ly, ch[0]->ly += ly;if (ch[1]) ch[1]->v += ly, ch[1]->m += ly, ch[1]->ly += ly;ly = 0;}}void Node::maintain() {s = 1, m = v;if (ch[0]) s += ch[0]->s, m = min(m, ch[0]->m);if (ch[1]) s += ch[1]->s, m = min(m, ch[1]->m);}Node* NewNode(int vv = 0, Node *f = NULL) {Node *ne = NN + (++totNN);ne->m = ne->v = vv; ne->fa = f;ne->ch[0] = ne->ch[1] = NULL;ne->s = 1; ne->rot = ne->ly = 0;return ne;}Node* Build(int l, int r, Node *f) {if (l > r) return NULL;int mid = (l + r) >> 1;Node *c = NewNode(a[mid], f);c->ch[0] = Build(l, mid - 1, c);c->ch[1] = Build(mid + 1, r, c);c->maintain();return c;}void Up(Node *u) {while (u) {u->maintain();u = u->fa;}}void Rot(Node *u) {if (u->fa == NULL) return;Node *f = u->fa, *ff = u->fa->fa;f->pushdown(); u->pushdown();int d = (u == f->ch[1]);Node *o = u->ch[d^1];f->ch[d] = o; f->fa = u;u->fa = ff; u->ch[d^1] = f;if (o) o->fa = f;if (ff) ff->ch[f == ff->ch[1]] = u;f->maintain(); u->maintain();}void splay(Node *u, Node *tag) {Node *f; int d, dd;while (u->fa != tag) {u->pushdown();if (u->fa->fa == tag) {Rot(u); break;}f = u->fa;d = u == f->ch[1], dd = f == f->fa->ch[1];if (d ^ dd) Rot(u);else Rot(f);Rot(u);}u->maintain();if (tag == NULL) root = u;}void RTO(int k, Node *tag) {Node *cur = root; int ls;while (cur) {cur->pushdown(); ls = 1;if (cur->ch[0]) ls += cur->ch[0]->s;if (ls == k) break;if (k < ls) cur = cur->ch[0];else {cur = cur->ch[1];k -= ls;}}if (cur) splay(cur, tag);}void Add(int x, int y, int val) {if (x > y) swap(x, y);RTO(x - 1, NULL); RTO(y + 1, root);Node* u = root->ch[1]->ch[0];u->v += val, u->ly += val, u->m += val;Up(u->fa); splay(u, NULL);}void Reverse(int x, int y) {if (x > y) swap(x, y);RTO(x - 1, NULL); RTO(y + 1, root);swap(root->ch[1]->ch[0]->ch[0], root->ch[1]->ch[0]->ch[1]);root->ch[1]->ch[0]->rot ^= 1;}void Insert(int k, int val) {RTO(k, NULL); RTO(k + 1, root);root->ch[1]->ch[0] = NewNode(val, root->ch[1]);Up(root->ch[1]);splay(root->ch[1]->ch[0], NULL);}void Delete(int k) {RTO(k - 1, NULL); RTO(k + 1, root);root->ch[1]->ch[0] = NULL; Up(root->ch[1]);splay(root->ch[1], NULL);}void Min(int x, int y) {if (x > y) swap(x, y);RTO(x - 1, NULL); RTO(y + 1, root);printf("%d\n", root->ch[1]->ch[0]->m);splay(root->ch[1]->ch[0], NULL); }void Revolve(int x, int y, int t) {if (x > y) swap(x, y);t %= (y - x + 1);t = (t + y - x + 1) % (y - x + 1);if (!t) return;RTO(y - t, NULL); RTO(y + 1, root);Node *u = root->ch[1]->ch[0];root->ch[1]->ch[0] = NULL; root->ch[1]->maintain(); root->maintain();RTO(x - 1, NULL); RTO(x, root);root->ch[1]->ch[0] = u;u->fa = root->ch[1];root->ch[1]->maintain(); root->maintain();}void Print(Node *u) {if (u == NULL) return;u->pushdown();Print(u->ch[0]);printf("%d ", u->v);Print(u->ch[1]);}//debugchar op[10];int main() {//freopen("T.in","r",stdin);//freopen("B.out","w",stdout);int n = getint(), x, y, d;for (int i = 1; i <= n; ++i) a[i] = getint();a[0] = a[n + 1] = ~0U>>1;root = Build(0, n + 1, NULL);n = getint();for (int i = n; i; --i) {scanf("%s",op);if (*op == 'A') {x = getint(); y = getint(); d = getint();Add(x + 1, y + 1, d);}else if (*op == 'I') {x = getint(); y = getint();Insert(x + 1, y);}else if (*op == 'D') Delete(getint() + 1);else if (*op == 'M') Min(getint() + 1, getint() + 1);else if (op[3] == 'E') {x = getint(); y = getint();Reverse(x + 1, y + 1);}else {x = getint(); y = getint(); d = getint();Revolve(x + 1, y + 1, d);}//Print(root);//puts("\n------");}}