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

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

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

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).

hi
bob

-

abca
accepted

ac

abacaba
abcdcba

abcba

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.

題目分析 ：

給出兩個字符串，刪除第二個字符串中的一些連續的字符，且要求剩下的字符是上面串的子串。

思路分析 ：

首先上來先想一個最暴力的思路，枚舉刪除字符串的長度，枚舉刪除的起點，再將剩下的字符串去匹配，雙指針，復雜度是 n^3 , 首先比較容易想到的一個地方優化，就是枚舉長度的時候我們可以去二分，復雜度為 logn ，但是僅憑這是不夠的，我們看一下字符串匹配那，是不是感覺匹配的過程中有很多匹配過程是重復的，那麽這裏肯定就是可以優化的，當刪除一個長度的字符串後，剩下的串一定是都要留下的，且可以看成一個前綴和一個後綴，那麽我們就可以預處理出全部可能的前綴以及後綴，新開兩個數組，記錄的每個位置對應在上面串中的哪個位置上，那麽總的復雜度就是 n*logn

代碼示例 ：

#define ll long long
const int maxn = 1e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

char pre[maxn], s[maxn];
int a[maxn], b[maxn];
int len1, len2;
int start = -1, len;

bool check(int lenn) {
    int p = lenn+1;
    int sign = 0;
    if (b[p]) {start = 1; len = lenn; sign = 1;}
    p = len2 - lenn;
    if (a[p]) {start = p+1; len = lenn; sign = 1;}
    //printf("****  %d %d\n", start, len); 
    for(int i = 2; i <= len2-lenn; i++){
        int ss = i-1, ee = i+lenn;
        if (!a[ss]) break;
        if (a[ss] < b[ee]) {start = i; len = lenn; sign = 1;}
    }
    if (sign) return true;
    else return false;
}

void fun() {
    int l = 0, r = len2;
    
    while(l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) r = mid - 1;
        else l = mid + 1;
        //printf("*****  %d  %d  %d  %d %d\n", start, len, mid, l, r);
        
    }
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    scanf("%s", pre+1);
    scanf("%s", s+1);
    len1 = strlen(pre+1);
    len2 = strlen(s + 1);
    int p = 1;
    for(int i = 1; i <= len2; i++){
        for(int j = p; j <= len1; j++){
            if (s[i] == pre[j]) {
                a[i] = j;
                p = j+1;
                break;
            }
        }
        if (!a[i]) break;
    }
    p = len1;
   for(int i = len2; i >= 1; i--){
        for(int j = p; j >= 1; j--){
            if (s[i] == pre[j]) {
                b[i] = j;
                p = j-1;
                break;
            }
        }
        if (!b[i]) break;
    }
    fun();
    //printf("%d %d\n", start, len);
    if (start == -1) {printf("-\n"); return 0;}
    for(int i = 1; i < start; i++) printf("%c", s[i]);
    for(int i = start+len; i <= len2; i++) printf("%c", s[i]);
    printf("\n");
    //for(int i = 1; i <= len2; i++) printf("%d ", b[i]);
    return 0;
}

二分 + 預處理前綴後綴 技巧題