POJ 1141 Brackets Sequence （区间DP）

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

()[()]

题意：给一串括号序列，按照合法括号的定义，添加若干括号，使得序列合法。

典型区间DP，设dp[i][j]为从i到j需要添加最少括号的数目。

dp[i][j] = max{ dp[i][k]+dp[k+1][j] }  (i<=k<j)

如果s[i] == s[j] , dp[i][j] 还要和dp[i+1][j-1]比较。 枚举顺序按照区间长度枚举。

因为要求输出合法序列，就要记录在原序列在哪些位置进行了增加，设c[i][j]为从i到j的 增加括号的位置，如果不需要增加，那么c[i][j] 赋为-1，打印时只需递归打印即可。

#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;typedef long long LL;const int MAX=0x3f3f3f3f;int n,c[105][105],dp[105][105];char s[105];void print(int i,int j) {    if( i>j ) return ;    if( i == j ) {        if(s[i] == '(' || s[i] == ')') printf("()");        else printf("[]");        return ;    }    if( c[i][j] > 0 ) {  // i到j存在增加括号的地方，位置为c[i][j]        print(i,c[i][j]);        print(c[i][j]+1,j);    } else {        if( s[i] == '(' ) {            printf("(");            print(i+1,j-1);            printf(")");        } else {            printf("[");            print(i+1,j-1);            printf("]");        }    }}void DP() {   //区间DP    for(int len=2;len<=n;len++)        for(int i=1;i<=n-len+1;i++) {            int j = i+len-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];                c[i][j] = k;  // 记录断开的位置            }            if( ( s[i] == '(' && s[j] == ')' || s[i] == '[' && s[j] == ']' ) && dp[i][j] > dp[i+1][j-1] ) {                dp[i][j] = dp[i+1][j-1];                c[i][j] = -1;  //i到j不需要断开，因为dp[i+1][j-1]的值更小，上面枚举的k位置都比这个大，所以不再断开            }        }}int main(){    scanf("%s",s+1);    n = strlen(s+1);    memset(c,-1,sizeof(c));    memset(dp,MAX,sizeof(c));    for(int i=1;i<=n;i++) dp[i][i] = 1, dp[i][i-1] = 0; //赋初值    DP();    print(1,n);    printf("\n");    return 0;}