【CODEFORCES】 C. Dreamoon and Strings

最新推荐文章于 2018-10-29 19:04:15 发布

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最新推荐文章于 2018-10-29 19:04:15 发布

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原文链接：http://www.cnblogs.com/llguanli/p/8514605.html

本文详细解析了一道关于字符串处理的问题，通过动态规划的方法，寻找在移除一定数量字符后，目标字符串中能匹配模式串的最大次数。并提供了一个完整的代码实现。

Dreamoon has a string s and a pattern string p. He first removes exactly x characters from s obtaining string s' as a result. Then he calculates $\text{[math]}$ that is defined as the maximal number of non-overlapping substrings equal to p that can be found in s'. He wants to make this number as big as possible.

More formally, let's define $\text{[math]}$ as maximum value of $\text{[math]}$ over all s' that can be obtained by removing exactly x characters from s. Dreamoon wants to know $\text{[math]}$ for all x from 0 to |s| where |s| denotes the length of string s.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 2 000).

The second line of the input contains the string p (1 ≤ |p| ≤ 500).

Both strings will only consist of lower case English letters.

Output

Print |s| + 1 space-separated integers in a single line representing the $\text{[math]}$ for all x from 0 to |s|.

Sample test(s)

input
aaaaa
aa

output
2 2 1 1 0 0

input
axbaxxb
ab

output
0 1 1 2 1 1 0 0

Note

For the first sample, the corresponding optimal values of s' after removal 0 through |s| = 5 characters from s are {"aaaaa", "aaaa","aaa", "aa", "a", ""}.

For the second sample, possible corresponding optimal values of s' are {"axbaxxb", "abaxxb", "axbab", "abab", "aba", "ab",

题解：这一题是DP。用D[i][j]表示从1至i。已经取走j个字符时的最大匹配串数。那么，第i位就有2种决策。取还是不取。而不取又分两种情况，一种是与前面的字符串没有能匹配的，一种是和前面的字符串有能匹配的。所以我们能够将状态转移方程表演示样例如以下：

D[i][j]=max(D[i-1][k-1],D[i-1][k]) //取还是不取

D[i][j]=max(D[i][j],D[k][j-(i-k-l2)]) //假设与前面有能匹配的情况

分析清楚了以后。能够发现这个问题跟最长不下降子序列事实上非常像。接下来就是编程实现了。要注意边界。D[i][j]的j不能大于i。并且不能为负数。

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

char s1[2005],s2[2005];
int l1,l2,j,d[2005][2005],f;

int mac(int i)
{
    int x=i,y=l2;
    while (x && y)
    {
        if (!y) break;
        if (s1[x-1]==s2[y-1]) {x--; y--;}
        else x--;
    }
    if (!y)
    {
        j=x;
        return 1;
    }
    else
    {
        j=0;
        return 0;
    }
}

int main()
{
    scanf("%s",s1);
    scanf("%s",s2);
    l1=strlen(s1);
    l2=strlen(s2);
    for (int i=1;i<=l1;i++)
    {
        f=mac(i);
        for (int k=0;k<=i;k++)
        {
            d[i][k]=max(d[i-1][k],d[i-1][k-1]);
            if (f && j>=k-i+j+l2 && k-i+j+l2>=0) d[i][k]=max(d[i][k],d[j][k-(i-j-l2)]+1);
        }
    }
    for (int i=0;i<=l1;i++)
        printf("%d ",d[l1][i]);
    return 0;
}

转载于:https://www.cnblogs.com/llguanli/p/8514605.html