字 符 串 字符串 字符串 | ( | ( | ( | ( | ) | ) |
---|---|---|---|---|---|---|
d p [ i ] dp[i] dp[i] | 1 | 2 | 3 | 4 | 3 | 2 |
m [ i ] m[i] m[i] | 1 | 2 | 2 | 2 | 2 | 2 |
字 符 串 字符串 字符串 | ( | ( | ) | ( | ) | ) |
---|---|---|---|---|---|---|
d p [ i ] dp[i] dp[i] | 1 | 2 | 1 | 2 | 1 |
字 符 串 字符串 字符串 | ) | ) | ) | ( | ( | ) |
---|---|---|---|---|---|---|
d p [ i ] dp[i] dp[i] | -1 | - 2 | - 3 | -2 | -1 | |
m [ i ] m[i] m[i] | -3 | -3 | -3 | -2 | -1 |
d p [ i ] = d p [ i − 1 ] + a [ i ] a [ i ] = { 1 s [ i ] = ( − 1 s [ i ] =    ) dp[i] = dp[i-1] + a[i] \quad a[i] = \begin{cases} 1 &\text {$s[i] = ($} \\ - 1 &\text {$s[i] = \;)$}\end{cases} dp[i]=dp[i−1]+a[i]a[i]={1−1s[i]=(s[i]=)
m [ i ] = m i n { m [ i ] , m [ i + 1 ] , ⋯   , m [ n − 1 ] } m[i] = min \lbrace m[i], m[i+1], \cdots, m[n-1] \rbrace m[i]=min{m[i],m[i+1],⋯,m[n−1]}
首先,我们可以知道,对于一个正确的括号匹配序列来说,
其 d p [ 0 ] = 1 , d p [ n − 1 ] = 0 , {dp[0] = 1, dp[n-1] = 0, } dp[0]=1,dp[n−1]=0, 并且最小值为0,假设其中有一个括号改变了,以 ) → ( 为 例 , a [ i ] = − 1 → 1 , 那 么 d p [ n − 1 ] = 2 ) \to( 为例,a[i] = -1 \to 1, 那么dp[n-1] = 2 )→(为例,a[i]=−1→1,那么dp[n−1]=2 显而易见,当一个错误括号序列,当且仅当改变其中一个括号就可以将其变为正确括号序列时,其 d p [ n − 1 ] = 2 ( o r − 2 ) {dp[n-1] = 2 (or -2) } dp[n−1]=2(or−2)
对于一个错误序列,我们对其的第i个括号修改,其 d p [ i ] , d p [ i + 1 ] , … , d p [ n − 1 ] 全 都 − 2 a [ i ] , dp[i], dp[i+1], \ldots, dp[n-1] 全都 -2a[i], dp[i],dp[i+1],…,dp[n−1]全都−2a[i],假设该操作正确,那么错误序列转化成正确序列,就要求 m [ i ] − 2 a [ i ] > = 0 { m[i] - 2a[i] >= 0 } m[i]−2a[i]>=0
详细代码如下:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
int awsl[maxn];
int main() {
int n;
scanf("%d", &n);
string s;
cin >> s;
vector<int> sum(n);
vector<int> m(n);
for (int i = 0; i < n; ++i) {
if (s[i] == '(') awsl[i] = 1;
else awsl[i] = -1;
}
sum[0] = awsl[0];
for(int i = 1; i<n; ++i) {
sum[i] = sum[i-1] + awsl[i];
}
m.back() = sum.back();
for(int i = n - 2; i >= 0; i--) {
m[i] = min(sum[i], m[i+1]);
}
int cnt = 0;
for(int i = 0; i < n; ++i) {
int v = 2*awsl[i];
if(m[i] - v >= 0 && sum.back() - v == 0) cnt++;
if(sum[i] < 0) break;
}
cout << cnt << '\n';
}