Repair the brckets FZU1987
这道题的关键就是知道一个修复一个括号序列的最小改变数
把'('变成1，')'变成-1，lmi为左起最小连续子段和(可以为空)，sum为序列的和，则答案为(lim+1)/2+(sum-lmi+1)/2
简单想了想，设i为lmi的结束下标，则有[1, i]所有的'('全部被匹配，[i+1, n]全部的‘)’都被匹配，如果不满足这个性质的话，则lmi还可以更小，这与假设相矛盾，所以-lmi就是[1,i]中没有被匹配的')'个数，而sum-lmi就是[i+1, n]没有被匹配的'('个数,所以只要分别修改[1,i]中的')'和[i+1]中的'('即可，由于原序列长度为偶数，所以|lmi|+|sum-lmi|一定是偶数,但
|lmi|和|sum-lmi|不一定是偶数,对于这种情况需要吧[1,i]中最后一个未匹配')'变为'(',[i+1,n]第一个未匹配'('变为')'，而|lmi|-1和|sum-lmi|-1又变成了偶数，综合这两种情况就可以得到上面的式子
剩下的就是splay的各种鬼畜操作了
这种单纯考察数据结构的题我都要调试N多遍才能通过，看来代码能力还是需要加强额
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>
#include <map>
#include <string>
#include <climits>
#include <set>
#include <string>
#include <sstream>
#include <utility>
#include <ctime>
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::istringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
const int MAXN(50010);
const int MAXM(10010);
const int MAXE(10010);
const int MAXH(19);
const int INFI((INT_MAX-1) >> 1);
const int MOD(10000);
const ULL BASE(31);
const LL LIM(10000000);
const int INV(-10000);
int arr[MAXN];
struct SPLAY_TREE
{
struct NODE
{
int num, size, flag1, flag2, flag3; //replace标记,reverse标记,inverse标记
int sum, lmi, rmi, lmx, rmx; //总和,左起连续最小和,右起连续最小和,左起连续最大和,右起连续最大和
NODE *fa;
NODE *ch[2];
};
NODE pool[MAXN];
NODE *root, *NIL, *rear;
inline void push_up(NODE *sour) //一定要注意NIL节点对push_up的影响
{
sour->size = sour->ch[0]->size+sour->ch[1]->size+1;
sour->sum = sour->ch[0]->sum+sour->ch[1]->sum+sour->num;
sour->lmi = min(sour->ch[0]->lmi, sour->ch[0]->sum+sour->num+sour->ch[1]->lmi);
sour->rmi = min(sour->ch[1]->rmi, sour->ch[1]->sum+sour->num+sour->ch[0]->rmi);
sour->lmx = max(sour->ch[0]->lmx, sour->ch[0]->sum+sour->num+sour->ch[1]->lmx);
sour->rmx = max(sour->ch[1]->rmx, sour->ch[1]->sum+sour->num+sour->ch[0]->rmx);
}
inline void updata_rep(NODE *sour, int v)
{
if(sour == NIL) return;
sour->num = v;
sour->sum = v*sour->size;
sour->rmi = sour->lmi = v > 0? 0: v*sour->size;
sour->lmx = sour->rmx = v > 0? v*sour->size: 0;
sour->flag1 = v;
sour->flag2 = 0;
sour->flag3 = 0;
}
inline void updata_rev(NODE *sour)
{
if(sour == NIL) return;
swap(sour->ch[0], sour->ch[1]);
swap(sour->lmi, sour->rmi);
swap(sour->lmx, sour->rmx);
sour->flag2 ^= 1;
}
inline void updata_inv(NODE *sour)
{
if(sour == NIL) return;
int temp = sour->lmi;
swap(sour->lmi, sour->lmx);
swap(sour->rmi, sour->rmx);
sour->lmi *= -1;
sour->lmx *= -1;
sour->rmi *= -1;
sour->rmx *= -1;
sour->num *= -1;
sour->sum *= -1;
if(sour->flag1)
sour->flag1 *= -1;
else
sour->flag3 ^= 1;
}
void push_down(NODE *sour) //和线段树的push_down一样，是对子节(注意不是本节点)进行更新
{
if(sour->flag1)
{
updata_rep(sour->ch[0], sour->flag1);
updata_rep(sour->ch[1], sour->flag1);
sour->flag1 = 0;
}
if(sour->flag2)
{
updata_rev(sour->ch[0]);
updata_rev(sour->ch[1]);
sour->flag2 = 0;
}
if(sour->flag3)
{
updata_inv(sour->ch[0]);
updata_inv(sour->ch[1]);
sour->flag3 = 0;
}
}
void initNIL()
{
NIL->ch[0] = NIL->ch[1] = NIL->fa = NIL;
NIL->lmi = NIL->rmi = NIL->lmx = NIL->rmx = NIL->sum = NIL->num = 0;
NIL->size = 0;
}
void init(int n)
{
NIL = pool;
initNIL();
rear = pool+1;
newnode(root, NIL, -100000); //插入无穷小
newnode(root->ch[1], root, 100000); //插入无穷大
build_tree(root->ch[1]->ch[0], root->ch[1], 1, n); //建树
push_up(root->ch[1]);
push_up(root);
}
void newnode(NODE *&sour, NODE *f, int num)
{
sour = rear++;
sour->num = sour->sum = num;
sour->size = 1;
sour->lmi = sour->rmi = num > 0? 0: num;
sour->lmx = sour->rmx = num > 0? num: 0;
sour->flag1 = sour->flag2 = sour->flag3 = 0;
sour->fa = f;
sour->ch[0] = sour->ch[1] = NIL;
}
void build_tree(NODE *&sour, NODE *f, int l, int r)
{
if(l > r)
return;
int m = (l+r) >> 1;
newnode(sour, f, arr[m]);
build_tree(sour->ch[0], sour, l, m-1);
build_tree(sour->ch[1], sour, m+1, r);
push_up(sour);
}
void rotate(NODE *sour, int flag)
{
NODE *f = sour->fa;
push_down(f);
push_down(sour);
f->ch[!flag] = sour->ch[flag];
sour->ch[flag]->fa = f;
sour->fa = f->fa;
if(f->fa != NIL)
f->fa->ch[f->fa->ch[1] == f] = sour;
sour->ch[flag] = f;
f->fa = sour;
push_up(f);
}
void splay(NODE *sour, NODE *goal)
{
push_down(sour);
while(sour->fa != goal)
{
if(sour->fa->fa == goal)
rotate(sour, sour->fa->ch[0] == sour);
else
{
NODE *f = sour->fa;
int flag = (f->fa->ch[0] == f);
if(f->ch[flag] == sour)
rotate(sour, !flag);
else
rotate(f, flag);
rotate(sour, flag);
}
}
push_up(sour);
if(goal == NIL)
root = sour;
}
NODE *select(NODE *sour, int r)
{
while(sour != NIL)
{
push_down(sour);
if(r == sour->ch[0]->size+1)
break;
if(r <= sour->ch[0]->size)
sour = sour->ch[0];
else
{
r -= sour->ch[0]->size+1;
sour = sour->ch[1];
}
}
return sour;
}
inline void pick(int pos1, int pos2)
{
NODE *tp = select(root, pos1);
splay(tp, NIL);
tp = select(root, pos2+2);
splay(tp, root);
}
void REPLACE(int pos1, int pos2, int value)
{
pick(pos1, pos2);
updata_rep(root->ch[1]->ch[0], value);
push_up(root->ch[1]);
push_up(root);
}
void SWAP(int pos1, int pos2)
{
pick(pos1, pos2);
updata_rev(root->ch[1]->ch[0]);
push_up(root->ch[1]);
push_up(root);
}
void INVERT(int pos1, int pos2)
{
pick(pos1, pos2);
updata_inv(root->ch[1]->ch[0]);
push_up(root->ch[1]);
push_up(root);
}
int QUERY(int pos1, int pos2)
{
pick(pos1, pos2);
NODE *tp = root->ch[1]->ch[0];
return (abs(tp->sum-tp->lmi)+1)/2+(abs(tp->lmi)+1)/2;
}
};
SPLAY_TREE spt;
char str[10];
int op1, op2;
char op3;
int main()
{
int TC, n_case(0);
scanf("%d", &TC);
while(TC--)
{
int n, m;
scanf("%d%d", &n, &m);
char temp;
for(int i = 1; i <= n; ++i)
{
scanf(" %c", &temp);
arr[i] = temp == '('? 1: -1;
}
spt.init(n);
printf("Case %d:\n", ++n_case);
for(int i = 0; i < m; ++i)
{
scanf("%s", str);
if(str[0] == 'R')
{
scanf("%d%d %c", &op1, &op2, &op3);
spt.REPLACE(op1, op2, op3 == '('? 1: -1);
}
else if(str[0] == 'S')
{
scanf("%d%d", &op1, &op2);
spt.SWAP(op1, op2);
}
else if(str[0] == 'I')
{
scanf("%d%d", &op1, &op2);
spt.INVERT(op1, op2);
}
else
{
scanf("%d%d", &op1, &op2);
printf("%d\n", spt.QUERY(op1, op2));
}
}
}
return 0;
}
又想了想，lmx和rmx是不必要维护的
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstring>
#include <stack>
#include <cctype>
#include <utility>
#include <map>
#include <string>
#include <climits>
#include <set>
#include <string>
#include <sstream>
#include <utility>
#include <ctime>
using std::priority_queue;
using std::vector;
using std::swap;
using std::stack;
using std::sort;
using std::max;
using std::min;
using std::pair;
using std::map;
using std::string;
using std::cin;
using std::cout;
using std::set;
using std::queue;
using std::string;
using std::istringstream;
using std::make_pair;
using std::getline;
using std::greater;
using std::endl;
using std::multimap;
using std::deque;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PAIR;
typedef multimap<int, int> MMAP;
const int MAXN(50010);
const int MAXM(10010);
const int MAXE(10010);
const int MAXH(19);
const int INFI((INT_MAX-1) >> 1);
const int MOD(10000);
const ULL BASE(31);
const LL LIM(10000000);
const int INV(-10000);
int arr[MAXN];
struct SPLAY_TREE
{
struct NODE
{
int num, size, flag1, flag2, flag3;
int sum, lmi, rmi; //区域和 ,左起连续最小和(可以0个元素), 右起连续最小和(可以0个元素)
NODE *fa;
NODE *ch[2];
};
NODE pool[MAXN];
NODE *root, *NIL, *rear;
inline void push_up(NODE *sour) //一定要注意NIL节点对push_up的影响
{
sour->size = sour->ch[0]->size+sour->ch[1]->size+1;
sour->sum = sour->ch[0]->sum+sour->ch[1]->sum+sour->num;
sour->lmi = min(sour->ch[0]->lmi, sour->ch[0]->sum+sour->num+sour->ch[1]->lmi);
sour->rmi = min(sour->ch[1]->rmi, sour->ch[1]->sum+sour->num+sour->ch[0]->rmi);
}
inline void updata_rep(NODE *sour, int v)
{
if(sour == NIL) return;
sour->num = v;
sour->sum = v*sour->size;
sour->rmi = sour->lmi = v > 0? 0: v*sour->size;
sour->flag1 = v;
sour->flag2 = 0;
sour->flag3 = 0;
}
inline void updata_rev(NODE *sour)
{
if(sour == NIL) return;
swap(sour->ch[0], sour->ch[1]);
swap(sour->lmi, sour->rmi);
sour->flag2 ^= 1;
}
inline void updata_inv(NODE *sour)
{
if(sour == NIL) return;
int temp = sour->lmi;
sour->lmi = -(sour->sum-sour->rmi);
sour->rmi = -(sour->sum-temp);
sour->sum *= -1;
sour->num *= -1;
if(sour->flag1)
sour->flag1 *= -1;
else
sour->flag3 ^= 1;
}
void push_down(NODE *sour) //和线段树的push_down一样，是对子节(注意不是本节点)进行更新
{
if(sour->flag1)
{
updata_rep(sour->ch[0], sour->flag1);
updata_rep(sour->ch[1], sour->flag1);
sour->flag1 = 0;
}
if(sour->flag2)
{
updata_rev(sour->ch[0]);
updata_rev(sour->ch[1]);
sour->flag2 = 0;
}
if(sour->flag3)
{
updata_inv(sour->ch[0]);
updata_inv(sour->ch[1]);
sour->flag3 = 0;
}
}
void initNIL()
{
NIL->ch[0] = NIL->ch[1] = NIL->fa = NIL;
NIL->lmi = NIL->rmi = NIL->sum = NIL->num = 0;
NIL->size = 0;
}
void init(int n)
{
NIL = pool;
initNIL();
rear = pool+1;
newnode(root, NIL, -100000); //插入无穷小
newnode(root->ch[1], root, 100000); //插入无穷大
build_tree(root->ch[1]->ch[0], root->ch[1], 1, n); //建树
push_up(root->ch[1]);
push_up(root);
}
void newnode(NODE *&sour, NODE *f, int num)
{
sour = rear++;
sour->num = sour->sum = num;
sour->size = 1;
sour->lmi = sour->rmi = num > 0? 0: num;
sour->flag1 = sour->flag2 = sour->flag3 = 0;
sour->fa = f;
sour->ch[0] = sour->ch[1] = NIL;
}
void build_tree(NODE *&sour, NODE *f, int l, int r)
{
if(l > r)
return;
int m = (l+r) >> 1;
newnode(sour, f, arr[m]);
build_tree(sour->ch[0], sour, l, m-1);
build_tree(sour->ch[1], sour, m+1, r);
push_up(sour);
}
void rotate(NODE *sour, int flag)
{
NODE *f = sour->fa;
push_down(f);
push_down(sour);
f->ch[!flag] = sour->ch[flag];
sour->ch[flag]->fa = f;
sour->fa = f->fa;
if(f->fa != NIL)
f->fa->ch[f->fa->ch[1] == f] = sour;
sour->ch[flag] = f;
f->fa = sour;
push_up(f);
}
void splay(NODE *sour, NODE *goal)
{
push_down(sour);
while(sour->fa != goal)
{
if(sour->fa->fa == goal)
rotate(sour, sour->fa->ch[0] == sour);
else
{
NODE *f = sour->fa;
int flag = (f->fa->ch[0] == f);
if(f->ch[flag] == sour)
rotate(sour, !flag);
else
rotate(f, flag);
rotate(sour, flag);
}
}
push_up(sour);
if(goal == NIL)
root = sour;
}
NODE *select(NODE *sour, int r)
{
while(sour != NIL)
{
push_down(sour);
if(r == sour->ch[0]->size+1)
break;
if(r <= sour->ch[0]->size)
sour = sour->ch[0];
else
{
r -= sour->ch[0]->size+1;
sour = sour->ch[1];
}
}
return sour;
}
inline void pick(int pos1, int pos2)
{
NODE *tp = select(root, pos1);
splay(tp, NIL);
tp = select(root, pos2+2);
splay(tp, root);
}
void REPLACE(int pos1, int pos2, int value)
{
pick(pos1, pos2);
updata_rep(root->ch[1]->ch[0], value);
push_up(root->ch[1]);
push_up(root);
}
void SWAP(int pos1, int pos2)
{
pick(pos1, pos2);
updata_rev(root->ch[1]->ch[0]);
push_up(root->ch[1]);
push_up(root);
}
void INVERT(int pos1, int pos2)
{
pick(pos1, pos2);
updata_inv(root->ch[1]->ch[0]);
push_up(root->ch[1]);
push_up(root);
}
int QUERY(int pos1, int pos2)
{
pick(pos1, pos2);
NODE *tp = root->ch[1]->ch[0];
return (abs(tp->sum-tp->lmi)+1)/2+(abs(tp->lmi)+1)/2;
}
};
SPLAY_TREE spt;
char str[10];
int op1, op2;
char op3;
int main()
{
int TC, n_case(0);
scanf("%d", &TC);
while(TC--)
{
int n, m;
scanf("%d%d", &n, &m);
char temp;
for(int i = 1; i <= n; ++i)
{
scanf(" %c", &temp);
arr[i] = temp == '('? 1: -1;
}
spt.init(n);
printf("Case %d:\n", ++n_case);
for(int i = 0; i < m; ++i)
{
scanf("%s", str);
if(str[0] == 'R')
{
scanf("%d%d %c", &op1, &op2, &op3);
spt.REPLACE(op1, op2, op3 == '('? 1: -1);
}
else if(str[0] == 'S')
{
scanf("%d%d", &op1, &op2);
spt.SWAP(op1, op2);
}
else if(str[0] == 'I')
{
scanf("%d%d", &op1, &op2);
spt.INVERT(op1, op2);
}
else
{
scanf("%d%d", &op1, &op2);
printf("%d\n", spt.QUERY(op1, op2));
}
}
}
return 0;
}