POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
				Special Judge

Description

Let us define a regular brackets sequence in the following way:

1. Empty sequence is a regular sequence.
2. If S is a regular sequence, then (S) and [S] are both regular sequences.
3. If A and B are regular sequences, then AB is a regular sequence.

For example, all of the following sequences of characters are regular brackets sequences:

(), [], (()), ([]), ()[], ()[()]

And all of the following character sequences are not:

(, [, ), )(, ([)], ([(]

Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

Input

The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output

Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

Sample Input

([(]

Sample Output

()[()]

Source

Northeastern Europe 2001

 

区间dp记录路径，结合了上两题的括号匹配+回文串修改。。

注意空行

#include<set>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 110;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,l,r) for(int i=l;i<=r;i++)
#define Down(i,r,l) for(int i=r;i>=l;i--)

struct State{
    int v,op;
}dp[N][N];

char st[N];
int n;

bool match(int x,int y){
    if(st[x]=='(') return (st[y]==')');
    if(st[x]=='[') return (st[y]==']');
    return false;
}
void DP(){
    n=strlen(st+1);
    For(i,n)
      For(j,n)
        if(j<i)  dp[i][j].v=0;
        else     dp[i][j].v=214748364;
    For(i,n) dp[i][i].v=1;
    Down(i,n-1,1)
      Rep(j,i,n){
          if(match(i,j))
            if(dp[i][j].v>dp[i+1][j-1].v){
                dp[i][j].v=dp[i+1][j-1].v;
                dp[i][j].op=0;
            }
          Rep(k,i,j-1)
            if(dp[i][k].v+dp[k+1][j].v<dp[i][j].v){
                dp[i][j].v=dp[i][k].v+dp[k+1][j].v;
                dp[i][j].op=k;
            }
      }
}

void Print(int l,int r){
    if(l>r) return;
    if(l==r){
        if(st[l]=='('||st[l]==')') printf("()");
        if(st[l]==']'||st[l]=='[') printf("[]");
        return;
    }
    if(dp[l][r].op){
        Print(l,dp[l][r].op);
        Print(dp[l][r].op+1,r);
    }
    else{
        printf("%c",st[l]);
        Print(l+1,r-1);
        printf("%c",st[r]);
    }
}

int main(){
    while(gets(st+1)){
        if(strlen(st+1)){
            DP();
            Print(1,n);
        }
        puts("");
    }
    return 0;
}