POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 24393 Accepted: 6885 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

思路：dp+dfs 搜索

写了很长的代码，本以为要调试好久呢，结果没有调试，没有注意到有输入空行的时候，用的scanf输入的，给为gets后ac了

#include <iostream>#include <cstring>#include <cmath>#include <algorithm>#include <cstdio>#include <cstdlib>#define N 110#define INF 0x7fffffusing namespace std;char s1[N];char ans[2*N],temp[N];int dp[N][N];struct num{    char c,next;}a[N*4],b[N*4];int back[N],front[N],Top1,Top2,Top3;int main(){    //freopen("data.txt","r",stdin);    void dfs(int l,int r);    while(gets(s1))    {        memset(dp,0,sizeof(dp));        int l = strlen(s1);        for(int i=l-1;i>=0;i--)        {            s1[i+1] = s1[i];        }        for(int i=1;i<=l;i++)        {            for(int j=i;j>=1;j--)            {                if(s1[i]=='['||s1[i]=='(')                {                   if(j==i)                   {                       dp[j][i] = 1;                   }else                   {                       dp[j][i] = dp[j][i-1]+1;                   }                   continue;                }                if(j==i)                {                    dp[j][i] = 1;                    continue;                }                int Min = INF;                for(int u=i;u>=j;u--)                {                    int s2=0,s3=0;                    if((s1[u]=='['&&s1[i]==']')||(s1[u]=='('&&s1[i]==')'))                    {                        if(j<=u-1)                        {                            s2+= dp[j][u-1];                        }                        if(u+1<=i-1)                        {                            s2+=dp[u+1][i-1];                        }                        Min = min(Min,s2);                    }                    if(j<=u-1)                    {                        s3+=(dp[j][u-1]);                    }                    if(u<=i-1)                    {                        s3+=dp[u][i-1];                    }                    s3+=1;                    Min = min(Min,s3);                }                dp[j][i] = Min;            }        }        memset(back,-1,sizeof(back));        memset(front,-1,sizeof(front));        Top1 = 0;        Top2 = 0;        dfs(1,l);        Top3 = 0;        for(int i=1;i<=l+1;i++)        {            int Top4 = 0;            if(i!=1)            {                Top4 = 0;                for(int j=back[i-1];j!=-1;j=a[j].next)                {                    temp[Top4++] = a[j].c;                }            }            if(i!=l+1)            {                for(int j=front[i];j!=-1;j=b[j].next)                {                   temp[Top4++] = b[j].c;                }            }            for(int j=Top4-1;j>=0;j--)            {                ans[Top3++] = temp[j];            }            if(i!=l+1)            {                ans[Top3++] = s1[i];            }        }        ans[Top3] = '\0';        printf("%s\n",ans);    }    return 0;}void addeage1(int x,char c){    a[Top1].c = c;    a[Top1].next = back[x];    back[x] = Top1++;}void addeage2(int x,char c){    b[Top2].c = c;    b[Top2].next = front[x];    front[x] = Top2++;}void dfs(int l,int r){    if(s1[r]=='('||s1[r]=='[')    {        if(s1[r]=='(')        {            addeage1(r,')');        }else        {            addeage1(r,']');        }        if(l<r)        {            dfs(l,r-1);        }    }else    {        if(l==r)        {            if(s1[r]==')')            {                addeage2(l,'(');            }else            {                addeage2(l,'[');            }            return ;        }        int ll,lr,rl,rr;        int k1=0,k2=0;        for(int i=l;i<=r;i++)        {            int sum = 0;            k1 = k2 = 0;            if(l<=i-1)            {                sum += dp[l][i-1];                k1 = 1;                ll = l;                lr = i-1;            }            if(i<=r-1)            {                sum+=dp[i][r-1];                k2 = 1;                rl = i;                rr = r-1;            }            sum+=1;            if(sum==dp[l][r])            {                if(s1[r]==')')                {                    addeage2(i,'(');                }else                {                    addeage2(i,'[');                }                break;            }            k1 = k2 = 0;            if((s1[i]=='('&&s1[r]==')')||(s1[i]=='['&&s1[r]==']'))            {                sum = 0;                if(l<=i-1)                {                    sum+=dp[l][i-1];                    k1 = 1;                    ll = l;                    rl = i-1;                }                if(i+1<=r-1)                {                    sum+=dp[i+1][r-1];                    k2 = 1;                    rl = i+1;                    rr = r-1;                }                if(sum==dp[l][r])                {                    break;                }            }        }        if(k1)        {            dfs(ll,lr);        }        if(k2)        {            dfs(rl,rr);        }    }}