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poj 1141 Brackets Sequence（区间dp）

2015-06-17 21:54 477 查看

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Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27063		Accepted: 7654		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

()[()]

给出只有括号的字符串求出最短的字符串能使所有的括号都能匹配

借鉴别人的思路：

思路：dp[i][j]表示从区间i到区间j使其所以括号匹配需要补上的最少括号数，那么当出现一个括号时，首先考虑它不与后面匹配的情况，那么需要加一个相对应的括号，让之匹配dp[i][j]=dp[i+1][j]+1;

然后再考虑，若是后面有括号可以让它匹配的情况，那么假设i<k<=j，当s[i]=='('&&s[k]==')'的时候，考虑动态转移，dp[i][j]=dp[i+1][k-1]+dp[k][j]-1

为什么这个动态方程减1呢，因为我将与之匹配的那个括号重新计算了一次，当s[k]==')'的时候，在计算dp[k][k]的时候，我的状态转移已经把这个括号自动匹配了一次，所以要减去这次匹配的........

然后就是记录路径了，开一个二维数组a[i][j]，当a[i][j]==-1的时候，表示dp[i][j]这个状态是从dp[i+1][j]推导过来的，当a[i][j]>0的时候，表示dp[i][j]是从dp[i+1][a[i][j]-1]以及dp[k][j]这两个状态推导过来的，那么注意到当a[i][j]！=-1的时候，就正好表示s[i]与s[a[i][j]]匹配，说明在第i个括号这个地方只需要输出它自己本身，其他的，若是a[i][j]==-1的，都需要输出它自身以及与它自身匹配的括号.........

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-8
#define MOD 10009
#define MAXN 10010
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char c = getchar();
    while (c < '0' || c > '9') c = getchar();
    int x = 0;
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

char ch[110];
int dp[110][110];
int pos[110][110],vis[110];

void visit(int s,int t)
{
    if(s>=t) return;
    if(pos[s][t]==-1)
        visit(s+1,t);
    if(pos[s][t]>0)
    {
        vis[s]=1;
        vis[pos[s][t]]=1;
        visit(s,pos[s][t]-1);
        visit(pos[s][t],t);
    }
}

int main()
{
//    fread;
    while(gets(ch))
    {
        int len=strlen(ch);
        len--;
//        cout<<"len  "<<len<<endl;
        MEM(vis,0); MEM(vis,0); MEM(pos,-1);
        for(int i=0;i<=len;i++)
            dp[i][i]=1;
        for(int i=len-1;i>=0;i--)
        {
            for(int j=i+1;j<=len;j++)
            {
                dp[i][j]=dp[i+1][j]+1;
                pos[i][j]=-1;
                for(int k=i+1;k<=j;k++)
                    if(ch[i]=='('&&ch[k]==')'||ch[i]=='['&&ch[k]==']')
                {
                    if(dp[i][j]>dp[i+1][k-1]+dp[k][j]-1)
                    {
                        dp[i][j]=dp[i+1][k-1]+dp[k][j]-1;
                        pos[i][j]=k;
                    }
                }
            }
        }
        visit(0,len);
        for(int i=0;i<=len;i++)
        {
            if(vis[i])
                printf("%c",ch[i]);
            else
            {
                if(ch[i]=='('||ch[i]==')')
                    printf("()");
                else printf("[]");
            }
        }
        puts("");
    }
    return 0;
}

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