poj 1141 Brackets Sequence（区间dp）

题目链接

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 26254 Accepted: 7394 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

题意：给一个只包含括号的串，问最少添加多少个串让该括号串合法，输出合法的括号串。

题解：我们考虑让括号串的第一括号匹配，那么，这个括号要么与已有的括号匹配，要么在某个位置添加一个与它匹配的括号。

我们可以用dp[l][r]表示让区间[l,r] 的括号串合法最少要添加多少个括号，转移就是：

1，如果第一个括号为')' 或 ']' ，显然只能在前面添加一个’(‘ 或']' 。

dp[l][r]=dp[l+1][r]+1;

2，如果第一个括号与第i个括号匹配，那么区间[l+1,i-1]和区间[i+1,r]的括号没有相互影响

dp[l][r]=min(dp[l+1][i-1]+dp[i+1][r])

3，如果在位置i和i+1之间添加一个括号与第一个括号匹配，那么区间[l+1,i]和区间[i+1,r] 之间没有影响

dp[l][r]=min(dp[l+1][i]+dp[i+1][r]+1)

在dp转移的过程中记录路径，最后输出方案即可。

代码如下：（注意由于空字符串也为合法串，所以输入的字符串有可能为空）

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<string>#include<math.h>#define nn 110#define inff 0x7fffffff#define eps 1e-8typedef long long LL;const LL inf64=LL(inff)*inff;using namespace std;int n;char s[nn];int dp[nn][nn];int lj[nn][nn];int dfs(int l,int r){    if(dp[l][r]!=-1)        return dp[l][r];    if(l>r)        return 0;    dp[l][r]=inff;    if(s[l]==')')    {        dp[l][r]=dfs(l+1,r)+1;        lj[l][r]=0;        return dp[l][r];    }    else if(s[l]==']')    {        dp[l][r]=dfs(l+1,r)+1;        lj[l][r]=0;        return dp[l][r];    }    int i;    for(i=2*l+2;i<=2*r+2;i++)    {        if(i%2)        {            if((s[l]=='('&&s[i/2]==')')||(s[l]=='['&&s[i/2]==']'))            {                if(dfs(l+1,i/2-1)+dfs(i/2+1,r)<dp[l][r])                {                    dp[l][r]=dfs(l+1,i/2-1)+dfs(i/2+1,r);                    lj[l][r]=i;                }            }        }        else        {            if((dfs(l+1,i/2-1)+dfs(i/2,r)+1)<dp[l][r])            {                dp[l][r]=dfs(l+1,i/2-1)+dfs(i/2,r)+1;                lj[l][r]=i;            }        }    }    return dp[l][r];}void go(int l,int r){    string re="";    if(l>r)        return ;    if(lj[l][r]==0)    {        if(s[l]==')')            printf("()");        else            printf("[]");        go(l+1,r);        return ;    }    if(lj[l][r]%2)    {        if(s[l]=='(')        {            printf("(");            go(l+1,lj[l][r]/2-1);            printf(")");            go(lj[l][r]/2+1,r);        }        else        {            putchar('[');            go(l+1,lj[l][r]/2-1);            putchar(']');            go(lj[l][r]/2+1,r);        }    }    else    {        if(s[l]=='(')        {            putchar('(');            go(l+1,lj[l][r]/2-1);            putchar(')');            go(lj[l][r]/2,r);        }        else        {            putchar('[');            go(l+1,lj[l][r]/2-1);            putchar(']');            go(lj[l][r]/2,r);        }    }}int main(){    int i;    while(gets(s))    {        int len=strlen(s);        memset(dp,-1,sizeof(dp));        dfs(0,len-1);        //cout<<dfs(0,len-1)<<endl;        go(0,len-1);        puts("");    }    return 0;}