C. Bracket Sequence（栈模拟 | DP）
作者：互联网
A bracket sequence is a string, containing only characters "(", ")", "[" and "]".
A correct bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()[]", "([])" are correct (the resulting expressions are: "(1)+[1]", "([1+1]+1)"), and "](" and "[" are not. The empty string is a correct bracket sequence by definition.
A substring s[l... r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2... s|s| (where |s| is the length of string s) is the string slsl + 1... sr. The empty string is a substring of any string by definition.
You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.
Input
The first and the only line contains the bracket sequence as a string, consisting only of characters "(", ")", "[" and "]". It is guaranteed that the string is non-empty and its length doesn't exceed 105 characters.
Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples
input
Copy
([])
output
Copy
1
([])
input
Copy
(((
output
Copy
0
栈模拟
匹配 [ ] 和 （ ） ，将" [ " 或"(" 入栈 (是把它在字符串中的相应位置入栈，方便分段查询），若匹配到相应的括号，则出栈，否则则压入。最后若栈中还有元素，即为区间断点，假设有m个元素，则分成m+1个区间。下边是贴的别人的代码。
AC Code：
#include <cstdio>
#include <map>
#include <stack>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif // ONLINE_JUDGE
map<char,int> m;
m['('] = 1;//匹配方便
m[')'] = -1;
m['['] = 2;
m[']'] = -2;
char str[100000+22];
int l;
int ans;
while (~scanf ("%s",str))
{
l = strlen(str);
stack<int> s;
ans = 0;
for (int i = 0 ; i < l ; i++)
{
if (m[str[i]] == 1 || m[str[i]] == 2)//将左括号入栈
s.push(i);
else if (s.empty())
s.push(i);
else if (m[str[i]] + m[str[s.top()]] == 0) //可以配对，弹出
s.pop();
else //不能配对，压入
s.push(i);
}
if (s.empty()) //全部完成匹配
{
for (int i = 0 ; i < l ; i++)
if (m[str[i]] == 2)
ans++;
printf ("%d\n%s\n",ans,str);
}
else//分区间查询
{
int st,endd;
int ans_st,ans_endd; //满足条件的开始于结束位置
int ant;
s.push(l);
while (s.size() != 1)
{
endd = s.top() - 1;
s.pop();
st = s.top() + 1;
ant = 0;
for (int i = st ; i <= endd ; i++)
if (m[str[i]] == 2)
ant++;
if (ant > ans)
{
ans = ant;
ans_st = st;
ans_endd = endd;
}
}
st = 0;
endd = s.top() - 1;
ant = 0;
for (int i = st ; i <= endd ; i++)
if (m[str[i]] == 2)
ant++;
if (ant > ans)
{
ans = ant;
ans_st = st;
ans_endd = endd;
}
if (ans)
{
printf ("%d\n",ans);
for (int i = ans_st ; i <= ans_endd ; i++)
printf ("%c",str[i]);
printf ("\n");
}
else
printf ("0\n\n");
}
}
return 0;
}
然后我自己写一遍
AC Code：
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<bitset>
#include<cctype>
#define LL long long
#define maxn (LL)1e5
#define INF 0x3f3f3f3f
const double eps = 0.0000001;
using namespace std;
inline int sgn(double x) {
return (x > eps) - (x < -eps);
}
map<char,int> st;
char c[100005];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
#ifndef ONLINE_JUDGE
//freopen("input.txt","r",stdin);
#endif // ONLINE_JUDGE
st['('] = 1;
st[')'] = -1;
st['['] = 2;
st[']'] = -2;
cin>>(c+1);
int l = strlen(c+1);
stack<int> s;
int k = 0;
for(int i =1;i<=l;++i)
{
if(c[i]=='('||c[i]=='[') {s.push(i);if(c[i]=='['){k++;}}
else if(s.empty()) s.push(i);
else if(st[c[i]]+st[c[s.top()]]==0) s.pop();
else s.push(i);
}
if(s.empty()) {
cout<<k<<endl;
cout<<(c+1);
return 0;
}
else {
s.push(l+1);
int star, end;
int Left,right;
int ans = 0;
while(s.size()!=1)
{
right = s.top()-1;
s.pop();
Left = s.top()+1;
int res = 0;
for(int i = Left;i<=right;++i)
{
if(c[i] == '[') res++;
}
if(res>ans){
ans = res;
star = Left;
end = right;
}
}
right = s.top()-1;
int res = 0;
for(int i = 1;i<=right;++i)
{
if(c[i] == '[') res++;
}
if(ans<res) {
ans = res;
star = 1;
end = right;
}
if(ans == 0) cout<<0<<endl;
else {
cout<<ans<<endl;
for(int i = star;i<=end;++i)
cout<<c[i];
return 0;
}
}
}
标签：string,Sequence,int,sequence,st,Bracket,ans,include,DP 来源： https://blog.csdn.net/qq_43408238/article/details/97492175