Codeforces Gym 100803G Flipping Parentheses 线段树+二分

Flipping Parentheses

题目连接：

http://codeforces.com/gym/100803/attachments

Description

A string consisting only of parentheses ‘(’ and ‘)’ is called balanced if it is one of the following.

• A string “()” is balanced.

• Concatenation of two balanced strings are balanced.

• When a string s is balanced, so is the concatenation of three strings “(”, s, and “)” in this
order.

Note that the condition is stronger than merely the numbers of ‘(’ and ‘)’ are equal. For instance,
“())(()” is not balanced.

Your task is to keep a string in a balanced state, under a severe condition in which a cosmic ray
may flip the direction of parentheses.

You are initially given a balanced string. Each time the direction of a single parenthesis is
flipped, your program is notified the position of the changed character in the string. Then,
calculate and output the leftmost position that, if the parenthesis there is flipped, the whole
string gets back to the balanced state. After the string is balanced by changing the parenthesis
indicated by your program, next cosmic ray flips another parenthesis, and the steps are repeated
several times

Input

The input consists of a single test case formatted as follows.

The first line consists of two integers N and Q (2 ≤ N ≤ 300000, 1 ≤ Q ≤ 150000). The second
line is a string s of balanced parentheses with length N. Each of the following Q lines is an
integer qi (1 ≤ qi ≤ N) that indicates that the direction of the qi-th parenthesis is flipped.

Output

For each event qi

, output the position of the leftmost parenthesis you need to flip in order to
get back to the balanced state.
Note that each input flipping event qi
is applied to the string after the previous flip qi−1 and its
fix.

Sample Input

6 3

((()))

4

3

1

Sample Output

2

2

1

Hint

题意

给你一个平衡的括号序列，然后每次询问是让一个括号掉转方向，让你找到一个最左边的，改变方向之后能够使得序列平衡的括号

强制在线（就是每次询问完之后，保持修改

题解：

把(想成1,)想成-1,平衡的显然就是前缀和为0

对于(改成)的，我们就找到最左边的)改成(就好了

对于)改成(的，我们就找到前缀和到结尾的最小值，大于等于2的第一个括号就好了

为什么呢？

因为我们维护的是前缀和，(改成)很显然是要-2的，所以我们只需要找到前缀和到结尾的最小值大于2的就好了，这样修改之后，也保持了平衡

我们用线段树+二分来解决

n(logn+logn)的和nlognlogn 的都比较好想

代码

#include<bits/stdc++.h>
using namespace std;

typedef int SgTreeDataType;
struct treenode
{
    int L , R  ;
    SgTreeDataType sum , lazy;
    SgTreeDataType now;
    void updata(SgTreeDataType v)
    {
        sum += v;
        lazy += v;
    }
};

treenode tree[300005*4];

void push_down(int o)
{
    SgTreeDataType lazyval = tree[o].lazy;
	tree[2*o].updata(lazyval) ; tree[2*o+1].updata(lazyval);
	tree[o].lazy = 0;
}

void push_up(int o)
{
	tree[o].sum = min(tree[2*o].sum , tree[2*o+1].sum);
}

void build_tree(int L , int R , int o)
{
	tree[o].L = L , tree[o].R = R, tree[o].lazy = 0;
	tree[o].now = 1e9;
	tree[o].sum = 0;
	if (R > L)
	{
		int mid = (L+R) >> 1;
		build_tree(L,mid,o*2);
		build_tree(mid+1,R,o*2+1);
	}
}

void updata_pre(int QL,int QR,SgTreeDataType v,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR) tree[o].updata(v);
	else
	{
		push_down(o);
		int mid = (L+R)>>1;
		if (QL <= mid) updata_pre(QL,QR,v,o*2);
		if (QR >  mid) updata_pre(QL,QR,v,o*2+1);
		push_up(o);
	}
}

void updata_idx(int QL,int QR,SgTreeDataType v,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR)tree[o].now = v;
	else
	{
		int mid = (L+R)>>1;
		if (QL <= mid) updata_idx(QL,QR,v,o*2);
		if (QR >  mid) updata_idx(QL,QR,v,o*2+1);
		tree[o].now = min(tree[o*2].now , tree[o*2+1].now);
	}
}

SgTreeDataType query(int QL,int QR,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR) return tree[o].sum;
	else
	{
		push_down(o);
		int mid = (L+R)>>1;
		SgTreeDataType res = 1e9;
		if (QL <= mid) res = min(res,query(QL,QR,2*o));
		if (QR > mid) res = min(res,query(QL,QR,2*o+1));
		push_up(o);
		return res;
	}
}
SgTreeDataType query2(int QL,int QR,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR) return tree[o].now;
	else
	{
		int mid = (L+R)>>1;
		SgTreeDataType res = 1e9;
		if (QL <= mid) res = min(res,query2(QL,QR,2*o));
		if (QR > mid) res = min(res,query2(QL,QR,2*o+1));
		return res;
	}
}
char str[300005];
int sum[300005];

int main()
{
    int n,q;
    scanf("%d%d",&n,&q);
    scanf("%s",str+1);
    build_tree(1,n,1);
    for(int i=1;i<=n;i++)
    {
        if(str[i]=='(')
        {
            updata_pre(i,n,1,1);
            updata_idx(i,i,1,1);
        }
        else
        {
            updata_pre(i,n,-1,1);
            updata_idx(i,i,-1,1);
        }
    }

    while(q--)
    {
        int x;
        scanf("%d",&x);
        if(str[x]=='(')
        {
            str[x]=')';
            updata_idx(x,x,-1,1);
            updata_pre(x,n,-2,1);
            int l = 1,r = x;
            while(l<=r)
            {
                int mid = (l+r)/2;
                if(query2(1,mid,1)<0)r=mid-1;
                else l=mid+1;
            }
            printf("%d
",l);
            updata_idx(l,l,1,1);
            updata_pre(l,n,2,1);
            str[l]='(';
        }
        else if(str[x]==')')
        {
            str[x]='(';
            updata_idx(x,x,1,1);
            updata_pre(x,n,2,1);
            int l = 1,r = x;
            while(l<=r)
            {
                int mid = (l+r)/2;
                if(query(mid,x,1)>=2)r=mid-1;
                else l=mid+1;
            }
            printf("%d
",l);
            updata_idx(l,l,-1,1);
            updata_pre(l,n,-2,1);
            str[l]=')';
        }
        //printf("%s
",str+1);
    }
}