POJ 1141 Brackets Sequence 区间dp

Brackets Sequence

POJ 1141
给一个由()[]组成的字符串，求将其补成Brackets Sequence所需的最小字符数。
(), [], (()), ([]), ()[], ()[()] 都是合法的序列。
区间动规，外层枚举区间长度：$dp[i][j] = \min_{i<=k<j}(dp[i][k] + dp[k+1][j])$特别地，当 $s_i,s_j$=‘()’或$s_i,s_j$=‘[]’时，有$dp[i][j]= \min_{i<=k<j}(d[i+1][j-1], \ dp[i][k] + dp[k+1][j])$.
题目要求输出添括号之后的数列，用$\begin{aligned} path[i][j] = \left\{ \begin{array}{lr} -1& s_i与s_j括号匹配\\ k & dp[i][k] + dp[k+1][j]最小时的划分点 \end{array} \right. \end{aligned}$然后递归输出匹配后的字符串。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int N = 1e2+10;
typedef std::pair<int, int> pii;
string s;
int n;
int dp[N][N],path[N][N];
void dfs(int l,int r){
    if(l>r) return;
    if(l==r) {
        if(s[l]=='('||s[l]==')') cout<<"()";
        if(s[l]=='['||s[l]==']') cout<<"[]";
        return;
    }
    if(path[l][r]==-1) {
        cout<<s[l];
        dfs(l+1,r-1);
        cout<<s[r];
        return;
    }
    dfs(l,path[l][r]);
    dfs(path[l][r]+1,r);
}
inline void init(){
    cin>>s;
    n=s.size();
    memset(dp,0,sizeof dp);
    for(int i=0;i<n;i++) dp[i][i]=1;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    init();
    for (int len = 1; len < n; len++)
        for (int i = 0; i < n - len; i++) {
            int j = i + len;
            dp[i][j]=inf;
            if ((s[i] == '[' && s[j] == ']') || (s[i] == '(' && s[j] == ')')) {
                dp[i][j] = dp[i + 1][j - 1];
                path[i][j] = -1;
            }
            int  pos=-1;
            for (int k = i; k < j; k++)
                if (dp[i][j] > dp[i][k] + dp[k + 1][j]) {
                    dp[i][j] = dp[i][k] + dp[k + 1][j];
                    pos = k;
                }
            if(pos!=-1) path[i][j]=pos;
        }
    dfs(0, n - 1);
    cout<<"\n";
    //std::cout<<dp[0][n-1]<<endl;
    return 0;
}