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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\n", 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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Sample Input
Sample Output
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Source
BZOJ 2329: [HNOI2011]括号修复( splay )
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原文地址：http://www.cnblogs.com/JSZX11556/p/5020634.html
