E. Bring Balance （括号匹配问题+贪心+前缀和）(CF 794 d2)

E. Bring Balance
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alina has a bracket sequence s of length 2n, consisting of n opening brackets '(' and n closing brackets ')'. As she likes balance, she wants to turn this bracket sequence into a balanced bracket sequence.

In one operation, she can reverse any substring of s.

What's the smallest number of operations that she needs to turn s into a balanced bracket sequence? It can be shown that it's always possible in at most n operations.

As a reminder, a sequence of brackets is called balanced if one can turn it into a valid math expression by adding characters + and 1. For example, sequences (())(), (), and (()(())) are balanced, while )(, ((), and (()))( are not.

Input
The first line of the input contains a single integer t (1≤t≤2⋅104)  — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤n≤105).

The second line of each test case contains a string s of length 2n, consisting of n opening and n closing brackets.

The sum of n over all test cases doesn't exceed 2⋅105.

Output
For each test case, in the first line output a single integer k (0≤k≤n)  — the smallest number of operations required.

The i-th of the next k lines should contain two integers li,ri (1≤li≤ri≤2n), indicating that in the i-th operation, Alina will reverse the substring slsl+1…sr−1sr. Here the numeration starts from 1.

If there are multiple sequences of operations with the smallest length which transform the sequence into a balanced one, you can output any of them.

Example
inputCopy
3
2
(())
5
())((()))(
6
())((()))(()
outputCopy
0
2
3 4
9 10
1
2 11
Note
In the first test case, the string is already balanced.

In the second test case, the string will be transformed as follows: ())((()))( → ()()(()))( → ()()(())(), where the last string is balanced.

In the third test case, the string will be transformed to ((()))((())), which is balanced.

思路：

• 核心： 把 ‘（’ 弄成 1，  ‘)’弄成 -1；，在利用前缀和， 画出前缀和的图像，
• 前缀和为0的点就是刚刚匹配好， 出现<0的点 就代表这个数列不合规了
• reverse 这个 某一段 就是把趋势对调，- + - + 变成 + - + -， L-1的值和 R的,值还是不变 (注意不是R+1)（前缀和嘛）
• 于是对于起点-1和终点值一样的reverse 就是相当于前缀和图像向上面对翻
• 本题:
• 结论 1 , 最多2 次就可以把 这个数列搞合格 (本题 n个1和n个-1) ,n>>=1;
• reverse 1- x, x+1,n (x,前缀和最大的那个点)
• 贪心简要推: 1-X 先把 上升趋势给他搞上去,之后不管怎么变 都不会小于0,(Codeforce 有严格推论)
• X+1 - n, 把下降的尽量往后边放.
• 所以 只需要在判读 0,1步骤有没有
• 0,就直接判断就行了 arr[i]>=0;
• 1, 首先找到 L ,R 把所有的小于0的点包裹进去
• 然后让 L 变成 max(arr[1]-arr[L])的点 贪心思想: 让前面的值(起点)尽量大,为后面提供容错,下降趋势往后面衍生,
• R 也是选一个后面的最大值, 贪心思想: 让一个尽量大的上升趋势往前面排.
• 注意具体操作时的 节点问题 , L 和R 有点区别的. L要+1,(i=0开始,防止 l=1,还要+1,(L要=1)) (一开始就下降的时候)

#include <bits/stdc++.h>
using namespace std;
#define M 400005
#define ri register int 

template <class G> void read(G &x)
{
    x=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){f|=ch=='-';ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    x=f?-x:x;
    return ;
}


int t;
int n;
long long arr[M];
int main(){
    ios::sync_with_stdio(false);
  cin.tie(0);
    cin>>t;
    while(t--)
    {
        cin>>n;
        n*=2;
        string s;
        cin>>s;
        vector<int>p(n+1);
        int mx=1;
        for(ri i=0;i<s.length();i++)
        {
            if(s[i]=='(') p[i+1]=1;
            else p[i+1]=-1;
            arr[i+1]=arr[i]+p[i+1];
            if(arr[mx]<arr[i+1]) mx=i+1;
        }
        int flag=1;
        for(ri i=1;i<=n;i++)
        {
            if(arr[i]<0) 
            {
                flag=0;break;
            }
        }
        if(flag)
        {
            printf("0\n");
            continue;
        }
        int l=0,r=0;
        for(ri i=1;i<=n;i++)
        {
            if(arr[i]<0)
            {
                l=i; break; 
            }
        }
        for(ri i=n;i>=1;i--)
        {
            if(arr[i]<0)
            {
                r=i; break; 
            }
        }
        r++;
        int tmp=l;
        for(ri i=0;i<=tmp;i++)
        {
            if(arr[i]>arr[l]) l=i;
        }
        tmp=r;
        for(ri i=tmp+1;i<=n;i++)
        {
            if(arr[i]>arr[r]) r=i;
        }
        l++;
        reverse(s.begin()+l-1,s.begin()+r+1-1);
        flag=1;
        for(ri i=0;i<s.length();i++)
        {
            if(s[i]=='(') p[i+1]=1;
            else p[i+1]=-1;
            arr[i+1]=arr[i]+p[i+1];
            if(arr[i+1]<0) 
            {
                flag=0;break;
            }
        }
        if(flag)
        {
            printf("1\n");
            printf("%d %d\n",l,r);
            continue;
        }
        printf("2\n");
        printf("1 %d %d %d\n",mx,mx+1,n);
    }
    return 0;
    
} 

最后小声BB,他题目还要要求L,R尽量小嘛?????????

标签：794,前缀,sequence,CF,balanced,Bring,test,ri,string
来源： https://www.cnblogs.com/Lamboofhome/p/16347602.html