POJ1141 Brackets Sequence

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

正解：DP

解题报告：

　　DP题，乍一看我居然不会做，也是醉了。开始想用贪心水过，发现会gi烂。

　　详细博客：http://blog.csdn.net/lijiecsu/article/details/7589877

　　不详细说了，见代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN = 1011;
char ch[MAXN];
int l;
int f[MAXN][MAXN],c[MAXN][MAXN];

inline void output(int l,int r){
    if(l>r) return ;
    if(l==r) {
    if(ch[l]=='(' || ch[l]==')') printf("()");
    else printf("[]");
    }
    else{
    if(c[l][r]>=0) {
        output(l,c[l][r]);
        output(c[l][r]+1,r);
    }
    else{
        if(ch[l]=='(') {
        printf("(");
        output(l+1,r-1);
        printf(")");
        }
        else{
        printf("[");
        output(l+1,r-1);
        printf("]");
        }
    }
    }
}

inline void solve(){
    scanf("%s",ch);
    int len=strlen(ch);
    for(int i=0;i<len;i++) f[i][i]=1;
    for(int i=0;i<len;i++) for(int j=0;j<len;j++) c[i][j]=-1;
    for(int l=1;l<=len-1;l++) 
    for(int i=0;i+l<=len-1;i++){
        int j=i+l;
        int minl=f[i][i]+f[i+1][j];
        c[i][j]=i;
        for(int k=i+1;k<j;k++){
        if(minl>f[i][k]+f[k+1][j]) {
            minl=f[i][k]+f[k+1][j];
            c[i][j]=k;
        }
        }
        f[i][j]=minl;

        if(( ch[i]=='(' && ch[j]==')' ) || ( ch[i]=='[' && ch[j]==']' )) {
        if(f[i][j]>f[i+1][j-1]) {
            f[i][j]=f[i+1][j-1];
            c[i][j]=-1;
        }
        }
    }

    output(0,len-1);
    printf("
");
}

int main()
{
    solve();
    return 0;
}