BZOJ 2329: [HNOI2011]括号修复( splay )

把括号序列后一定是))))((((这种形式的..所以维护一个最大前缀和l, 最大后缀和r就可以了..答案就是(l+1)/2+(r+1)/2...用splay维护,O(NlogN). 其实还是挺好写的, 只是我傻X

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#include<cstdio>

#include<algorithm>

#include<cstring>

 

using namespace std;

 

const int maxn = 100009;

 

struct Node* Null;

 

struct Node {

Node *p, *ch[2];

int s, v;

int lmn, lmx, rmn, rmx, sm;

int tg;// tg 1) -1(

bool rev, inv;

inline void Clr() {

lmn = lmx = rmn = rmx = sm = s = v = tg = rev = inv = 0;

}

inline bool d() {

return this == p->ch[1];

}

inline void setc(Node* t, int c) {

ch[c] = t;

t->p = this;

}

inline void Tg(int v) {

tg = v;

rev = inv = false;

}

inline void Rev() {

rev ^= 1;

}

inline void Inv() {

inv ^= 1;

}

inline void pushDown() {

if(tg) {

v = tg;

for(int i = 0; i < 2; i++)

if(ch[i] != Null) ch[i]->Tg(tg);

tg = 0;

}

if(rev) {

swap(ch[0], ch[1]);

for(int i = 0; i < 2; i++)

if(ch[i] != Null) ch[i]->Rev();

rev = false;

}

if(inv) {

v = -v;

for(int i = 0; i < 2; i++)

if(ch[i] != Null) ch[i]->Inv();

inv = false;

}

for(int i = 0; i < 2; i++)

if(ch[i] != Null) ch[i]->upd();

}

inline void upd() {

lmx = max(ch[0]->lmx, ch[0]->sm + max(0, v + max(0, ch[1]->lmx)));

rmx = max(ch[1]->rmx, ch[1]->sm + max(0, v + max(0, ch[0]->rmx)));

lmn = min(ch[0]->lmn, ch[0]->sm + min(0, v + min(0, ch[1]->lmn)));

rmn = min(ch[1]->rmn, ch[1]->sm + min(0, v + min(0, ch[0]->rmn)));

sm = ch[0]->sm + v + ch[1]->sm;

s = ch[0]->s + 1 + ch[1]->s;

if(tg) {

sm = tg * s;

lmx = max(0, sm);

rmx = max(0, sm);

lmn = min(0, sm);

rmn = min(0, sm);

}

if(inv) {

swap(lmn, lmx); lmn = -lmn; lmx = -lmx;

swap(rmn, rmx); rmn = -rmn; rmx = -rmx;

sm = -sm;

}

if(rev) {

swap(lmx, rmx);

swap(lmn, rmn);

}

}

inline int Query() {

return (lmx + 1) / 2 + (-rmn + 1) / 2;

}

} *Root, pool[maxn], *pt;

 

void InitSplay() {

pt = pool;

Root = Null = pt++;

Null->Clr();

Null->setc(Null, 0);

Null->setc(Null, 1);

}

 

Node* NewNode(int v) {

pt->Clr();

pt->v = v;

pt->s = 1;

pt->p = Null;

return pt++;

}

 

void Rot(Node* t) {

Node* p = t->p;

p->pushDown();

t->pushDown();

int d = t->d();

p->p->setc(t, p->d());

p->setc(t->ch[d ^ 1], d);

t->setc(p, d ^ 1);

p->upd();

if(p == Root)

Root = t;

}

 

Node* Select(int k) {

for(Node* t = Root; ; ) {

t->pushDown();

int s = t->ch[0]->s;

if(k == s)

return t;

if(k > s)

k -= s + 1, t = t->ch[1];

else

t = t->ch[0];

}

}

 

void Splay(Node* t, Node* f = Null) {

for(Node* p = t->p; p != f; p = t->p) {

if(p->p != f)

t->d() != p->d() ? Rot(t) : Rot(p);

Rot(t);

}

t->upd();

}

 

Node* Range(int l, int r) {

Splay(Select(--l));

Splay(Select(++r), Root);

return Root->ch[1]->ch[0];

}

 

int N;

char s[maxn];

 

Node* Build(int l, int r) {

if(l >= r)

return Null;

int m = (l + r) >> 1;

Node* t = NewNode(s[m] != ')' ? -1 : 1);

t->setc(Build(l, m), 0);

t->setc(Build(m + 1, r), 1);

t->upd();

return t;

}

 

int main() {

InitSplay();

int m;

scanf("%d%d", &N, &m);

scanf("%s", s + 1);

Root = Build(0, N + 2);

while(m--) {

int l, r;

scanf("%s%d%d", s, &l, &r);

Node* t = Range(l, r);

if(s[0] == 'R') {

char c;

scanf(" %c", &c);

t->Tg(c != '(' ? 1 : -1);

} else if(s[0] == 'S') {

t->Rev();

} else if(s[0] == 'I') {

t->Inv();

} else

printf("%d ", t->Query());

Splay(t);

}

return 0;

}

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2329: [HNOI2011]括号修复

Time Limit: 40 Sec  Memory Limit: 128 MB
Submit: 798  Solved: 368
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Description

Input

Output

Sample Input

Sample Output

HINT

Source

Splay