poj1141（区间dp基础）

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 29414 Accepted: 8368 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

()[()]

题意：

给出只含有'('')''['']'这四种字符的字符串，要求让这个串匹配的最少需要添加的字符个数并且输出该字符串。

思路：

这题看起来比较水，可是自己想不是太好想，看了别人说要递归自己试了下还行，记忆化搜索写起来挺顺手，不过可惜这题刚开始把状态转移方程搞错了总是wa。

用dp[i][j]表示区间i到j之间的最少的需要匹配的字符对数，状态转移是：

如果x[i]和x[j]匹配，  dp[i][j]=min(dp[i+1][j-1]+1,dp[i][k]+dp[k+1][j]),

否则，  dp[i][j]=min(dp[i][k]+dp[k+1][j])

fen[i][j]表示区间i到j如果dp[i][j]=某一个dp[i][k]+dp[k+1][j],那么fen[i][j]=k表示区间i到j是从k这个地方分开的！而且这个fen[i][j]数组很必要，否则还得在输出的函数里要加上一大堆判断条件！

#include <iostream>#include <stdio.h>#include <stdlib.h>#include<string.h>#include<algorithm>#include<math.h>#include<queue>using namespace std;typedef long long ll;int dp[110][110];int fen[110][110];char x[110];int fdp(int a,int b){    if(dp[a][b]!=-1)return dp[a][b];    if(a==b)return dp[a][b]=1;    if(a>b)return dp[a][b]=0;    int min0=1e9;    if((x[a]=='('&&x[b]==')')||(x[a]=='['&&x[b]==']'))        min0=fdp(a+1,b-1)+1;    for(int k=a; k<b; k++)        if(min0>fdp(a,k)+fdp(k+1,b))            min0=fdp(a,k)+fdp(k+1,b),fen[a][b]=k;    return dp[a][b]=min0;}void prin(int a,int b){    if(a==b)    {        if(x[a]=='('||x[a]==')')            printf("()");        else printf("[]");        return;    }    if(a>b)        return;    if(fen[a][b]==-1)    {        if(x[a]=='(')        {            printf("(");            prin(a+1,b-1);            printf(")");        }        else        {            printf("[");            prin(a+1,b-1);            printf("]");        }        return ;    }    prin(a,fen[a][b]);    prin(fen[a][b]+1,b);}int main(){    scanf("%s",x);    memset(dp,-1,sizeof(dp));    memset(fen,-1,sizeof(fen));    int l=strlen(x);    fdp(0,l-1);    prin(0,l-1);    puts("");    return 0;}