UVA1626 DP经典

最新推荐文章于 2020-01-26 23:29:08 发布

原创最新推荐文章于 2020-01-26 23:29:08 发布 · 527 阅读

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#算法竞赛入门经典第二版 #dp #uva

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该博客探讨了如何解决UVA1626算法竞赛题目，这是一个关于找到最短的正规括号序列，使其包含给定字符序列作为子序列的问题。通过动态规划方法，利用状态转移方程求解，强调了理解题目描述和处理特殊情况（如空字符串）的重要性。

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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
a1a2 … an is called a subsequence of the string b1b2 … bm, if there exist such indices 1 ≤ i1 < i2 <
… < in ≤ m, that aj = bij
for all 1 ≤ j ≤ n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below.
This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a
single line without any other characters among them.
Output
For each test case, the output must follow the description below. The outputs of two consecutive cases
will be separated by a blank line.
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
1
([(]
Sample Output
()[()]

本题是算法竞赛入门经典里面的题目，当时看书的时候觉得自己看明白了，但是当test时再次遇到的时候还是没做出来，说明功底还是不够，经典的动态规划。状态转移方程
当s[i],s[j]能够匹配时,f(i,j)=f(i+1,j-1)。判断之后才有f(i,j)=min{f(i,k)+f(k+1,j) | i < k < j}有一点很重要，就是不管是否满足第一种状态，都要尝试第二种状态。

还有这道题对我来说最坑的一点就是上面加粗斜体的那句话。在输入之前会输入一个空行。还有可能会输入空字符串。

#include <cstdio>
#include <iostream>
using namespace std;
#include <algorithm>
#include <cstring>
int n;
char s[105];
int d[105][105];
int match(char a,char b)
{
    if((a=='('&&b==')')||(a=='['&&b==']'))
        return 1;
    return 0;
}
void dp()
{
    for(int i=0;i<n;i++)
    {
        d[i+1][i]=0;
        d[i][i]=1;
    }
    for(int i=n-2;i>=0;i--)
    {
        for(int j=i+1;j<n;j++)
        {
            d[i][j]=n;
            if(match(s[i],s[j])) d[i][j]=min(d[i][j],d[i+1][j-1]);
            for(int k=i;k<j;k++)
                d[i][j]=min(d[i][k]+d[k+1][j],d[i][j]);
        }
    }
}
void print(int i,int j)
{
    if(i>j) return;
    if(i==j)
    {
        if(s[i]=='('||s[i]==')')
            printf("()");
        else
            printf("[]");
            return;
    }
    if(match(s[i],s[j])&&d[i][j]==d[i+1][j-1])
    {
        putchar(s[i]);
        print(i+1,j-1);
        putchar(s[j]);
        return;
    }
    for(int k=i;k<j;k++)
    {
        if(d[i][j]==d[i][k]+d[k+1][j])
        {
            print(i,k);
            print(k+1,j);
            return ;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        memset(d,0,sizeof(d));
        fgets(s,105,stdin);
        fgets(s,105,stdin);
        n=strlen(s)-1;
        //cout<<n<<endl;
        dp();
        print(0,n-1);
        cout<<endl;
        if(t)
            cout<<endl;
    }
    return 0;

}