Codeforces 762C Two strings (前缀与后缀)

最新推荐文章于 2018-04-18 21:58:18 发布

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本文介绍了一种特殊的字符串匹配问题，即如何从字符串B中删除最少数量的连续字符，使得剩余部分成为字符串A的子序列。文章通过使用前缀和后缀数组记录匹配长度，并通过枚举确定最优解的方法来解决该问题。

You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of the characters from b and make it empty.

Subsequence of string s is any such string that can be obtained by erasing zero or more characters (not necessarily consecutive) from string s.

Input

The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters.

Output

On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.

If the answer consists of zero characters, output «-» (a minus sign).

Examples

input
hi
bob

output
-

input
abca
accepted

output
ac

input
abacaba
abcdcba

output
abcba

Note

In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.

In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b.

123

不得不说CF的题目真是又爱又恨啊, 第一遍读上去,感觉像公共子序列的问题, ,看第二个示例,又不是, 晕啊晕,

题目大意: 给你A 串 B 串问在 B串中删除x连续的串后, 剩下的连在一起是A的子序列,

注意: A 可以不连续, 只要B剩下的字符, A里面有就可以

用两个数组, 一个前缀,一个后缀, 记录 A串中 B的子序列长度, 然后枚举A 从中选取删除后缀减去前缀的部分;

选个最小的; 题目是最小连续的B子串

注意赋初值的问题以及后缀最大的要存最大值; 是为了删除串的作用

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

using namespace std;

typedef long long ll;
const int MAXN=1e5+155;
int pre[MAXN],suf[MAXN];//存前缀和后缀    相当于存 A的  对于B的子序列的长度


char a[MAXN],b[MAXN];
int main()
{
    while(~scanf("%s%s",a+1,b+1))
    {
        int len1=strlen(a+1);
        int len2=strlen(b+1);
        int last=0;
        // 不论是前缀还是后缀,  只要A 中有B的 某些串就可以 可以不连续,  例如A :abc   B: ac   B为A的子串
        for(int i=1;i<=len1;i++)  //赋初值
            pre[i]=len2,suf[i]=1;
        suf[len1+1] = len2+1;// 必须有 目的是为了删除子串的作用, pre[i-1]
        for(int i=1,j=1;i<=len1&&j<=len2;i++)  //这里是构造A的前缀
        {
            if(a[i]==b[j])// 相同记录前缀, 更新
            {
                pre[i]=j;
                last=j;
                j++;
            }
            else
                pre[i]=last;  //如果这个不相同，就等于前面的last
        }
        last=len2+1;
        for(int i=len1,j=len2;i>=1&&j>=1;i--)//构造A的后缀
        {
            if(a[i]==b[j])
            {
                suf[i]=j;
                last =j;
                j--;
            }
            else
                suf[i]=last;
        }
        int ans=len2+1, left=1,right=len2+1; //初始化答案为最大;
        for(int i=1;i<=len1+1;i++)//枚举所有的情况
        {
            if(pre[i-1]<suf[i]&&ans>suf[i]-pre[i-1])
            {
                ans=suf[i]-pre[i-1];
                left=pre[i-1];
                right=suf[i];
            }
        }
        if(ans== len2+1 )
        {
            cout<<"-"<<endl;
            continue;
        }
        for(int i=1;i<=len2;i++)// 输出 删除后的b
        {
            if(i<=left||i>=right)
                cout<<b[i];
        }
    }
    return 0;
}