POJ 1141 Brackets Sequence 区间dp

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
()[()]

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思路：

简单区间dp，对于一段区间，分三种情况讨论：
1、这个区间被分成多块。（regular sequence算是一块）
2、整个是一块，且两边分别保留一个字符。
3、整个是一块，且只保留一边的一个字符。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
int dp[105][105],dp_cut[105][105];
char str[205];
int DP(int l,int r) {
    if(l>r) return 0;
    if(dp[l][r]!=-1) return dp[l][r];
    if(r==l) {
        return dp[l][r]=2;
    }
    int ans = 1000;
    int cut;
    if( (str[l]=='(' && str[r]==')') || (str[l]=='['&&str[r]==']') ) {
        if(2+DP(l+1,r-1)2+DP(l+1,r-1);
            cut = -2;
        }
    }
    if( str[l]=='(' || str[l]=='[' ) {
        if(2+DP(l+1,r)2+DP(l+1,r);
            cut = -3;
        }
    }
    if( str[r]==')' || str[r]==']' ) {
        if(2+DP(l,r-1)2+DP(l,r-1);
            cut = -1;
        }
    }
    for(int m=l;mif(DP(l,m)+DP(m+1,r)1,r);
            cut = m;
        }
    }
    dp_cut[l][r] = cut;
    return dp[l][r]=ans;
}
void print_ret(int l,int r) {
    if(l>r) return;
    if(l==r) {
        if(str[l]=='['||str[l]==']') printf("[]");
        else if(str[l]=='('||str[l]==')') printf("()");
    }
    else if(dp_cut[l][r]==-2) {
        printf("%c",str[l]);
        print_ret(l+1,r-1);
        printf("%c",str[r]);
    }
    else if(dp_cut[l][r]==-3) {
        printf("%c",str[l]);
        print_ret(l+1,r);
        if(str[l]=='[') printf("]");
        else if(str[l]=='(') printf(")");
    }
    else if(dp_cut[l][r]==-1) {
        if(str[r]==']') printf("[");
        else if(str[r]==')') printf("(");
        print_ret(l,r-1);
        printf("%c",str[r]);
    }
    else {
        int m = dp_cut[l][r];
        print_ret(l,m);
        print_ret(m+1,r);
    }
}
int main() {
    while(gets(str)!=NULL) {
        memset(dp,-1,sizeof(dp));
        int len = strlen(str);
        DP(0,len-1);
        print_ret(0,len-1);
        puts("");
    }
}