POJ 1141(Bracket Sequence)-区间DP

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 23158 Accepted: 6508 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

 

 

经典区间DP。。。

/************************************************* author:crazy_石头* Pro:POJ 1141 Bracket Sequence* algorithm:区间DP* Time:110ms* Judge Status:Accepted* 解题思路：* dp[i][j]表示i~j间最少添加的字符能使所有括号匹配* 那么当i和j匹配时，i~j的匹配数(dp[i][j])等于i+1~j-1* 间的匹配数，即dp[i][j]=dp[i+1][j-1];* 对于一般情况下，dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]* 不断枚举k即可；找出最小值，用path记录路径，path=-1* 表示i~j间括号得到匹配，输出递归输出即可；**************************************************/#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>using namespace std;#define LL long long#define rep(i,h,n) for(int i=(h);i<=(n);i++)#define ms(a,b) memset((a),(b),sizeof(a))#define INF 1<<29const int maxn=1500+5;const int maxm=32000+5;const LL mod=10000007 ;int dp[maxn][maxn];//i，j间最少添加几个字符串才匹配；int path[maxn][maxn];char str[maxn];inline bool match(int i,int j){    if((str[i]=='['&&str[j]==']')||(str[i]=='('&&str[j]==')'))    return true;    return false;}inline void output(int i,int j){    if(i>j)return;    if(i==j)    {        if(str[i]=='['||str[i]==']')printf("[]");        else printf("()");        return ;    }    if(path[i][j]==-1)//path[i][j]==-1时表示匹配；    {        printf("%c",str[i]);        output(i+1,j-1);        printf("%c",str[j]);        return ;    }    output(i,path[i][j]);    output(path[i][j]+1,j);}inline void solve(){    int len=strlen(str);    rep(i,0,len-1)    dp[i][i]=1;    for(int i=1;i<len;i++)    {        for(int j=0;j+i<len;j++)        {            int t=j+i;            dp[j][t]=INF;            if(match(j,t))            {                dp[j][t]=dp[j+1][t-1];                path[j][t]=-1;            }            for(int k=j;k<t;k++)            {                if(dp[j][t]>dp[j][k]+dp[k+1][t])                {                    dp[j][t]=dp[j][k]+dp[k+1][t];                    path[j][t]=k;                }            }        }    }}int main(){    while(gets(str))    {        ms(dp,0);        ms(path,0);        int len=strlen(str)-1;        solve();        output(0,len);        printf("\n");    }    return 0;}