poj 1141 Brackets Sequence dp

最新推荐文章于 2017-11-22 18:30:48 发布

joafelnglfds

最新推荐文章于 2017-11-22 18:30:48 发布

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分类专栏：动态规划

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本文探讨了如何找出包含特定字符序列的最短合法括号序列。通过动态规划方法，文章详细介绍了两种实现方式——自底向上和自顶向下的递归方法，并提供了完整的代码示例。

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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

()[()]

题目分析：

dp要找到子问题的所在。我自己看的别人的博客才知道怎么办；dp[i][j]表示i,j之间最少需要增加的符号个数。用path记录路径。状态转移dp[i][j]=min(dp[i][k]+dp[k+1][j]);意思是从哪个点分开的话最优。当str[i]==str[j]时dp[i][j]=dp[i+1][j-1];

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define eps 1e-9
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define M 10000+10
using namespace std;
int dp[200][200],path[200][200];
int n;
char str[200];
void print(int i,int j)
{
    if(i>j)  //前一步出现了已经匹配的状态
        return;
    if(i==j)
    {
        if(str[i]==')'||str[i]=='(')
            printf("()");
        else if(str[i]=='['||str[i]==']')
            printf("[]");
    }
    else if(path[i][j]==-1)
    {
        printf("%c",str[i]);
        print(i+1,j-1);
        printf("%c",str[j]);
    }
    else
    {
        int k=path[i][j];
        print(i,k);
        print(k+1,j);
    }
}
int main()
{

    int i,j,k;

    while(gets(str))
    {
        n=strlen(str);
        if(n==0)  //空串也是符合要求的
        {
            printf("\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        int len;
        for(len=1;lendp[j+1][k-1])
            {
                dp[j][k]=dp[j+1][k-1];
                path[j][k]=-1;  //表示不需要另外加括号就已经匹配
            }
            for(i=j;idp[j][i]+dp[i+1][k])
            {
                dp[j][k]=dp[j][i]+dp[i+1][k];
                path[j][k]=i;
            }
        }
        print(0,n-1);
        printf("\n");
    }

    return 0;
}
还有一种就是自顶向下，记忆化搜索。
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define ll long long
#define eps 1e-9
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const int M=5e4+10;
using namespace std;
int dp[200][200],path[200][200],vis[200][200];
int n;
char str[200];
int solve(int i,int j)
{
    int k;

    if(vis[i][j])
        return dp[i][j];
    int ans=INF;
    if(i>j)
        return 0;
    else if(i==j)
        return 1;
    if(str[i]=='('&&str[j]==')'||str[i]=='['&&str[j]==']')
    {
        int t1=solve(i+1,j-1);
        if(ans>t1)
        {
            ans=t1;
            path[i][j]=-1;
        }
    }

    for(k=i;kt1+t2)
        {
            ans=t1+t2;
            path[i][j]=k;
        }
    }


    dp[i][j]=ans;
    vis[i][j]=1;

    return ans;
}
void print(int i,int j)
{
    if(i>j)
        return;
    if(i==j)
    {
        if(str[i]==')'||str[i]=='(')
            printf("()");
        else if(str[i]=='['||str[i]==']')
            printf("[]");
    }
    else if(path[i][j]==-1)
    {
        printf("%c",str[i]);
        print(i+1,j-1);
        printf("%c",str[j]);
    }
    else
    {
        int k=path[i][j];
        print(i,k);
        print(k+1,j);
    }
}
int main()
{
    while(gets(str))
    {
        n=strlen(str);
        if(n==0)
        {
            printf("\n");
            continue;
        }
        memset(vis,0,sizeof(vis));
        solve(0,n-1);
        print(0,n-1);
        printf("\n");
    }
}