[Educational Codeforces Round 17 C (762C)] Two strings

題意

給定兩個串a、b，要求在b中刪掉一個子串，使得b串成爲a串的子序列

題解

從b串左邊開始匹配a串左邊，記錄下來匹配到的位置，知道匹配結束
再從b串右邊開始匹配a串右邊，記錄下來匹配到的位置。
。。。。然而這需要開兩個數組。。。否則就像我一樣gg了

代碼

/// by ztx

// blog.csdn.net/hzoi_ztx

#define  maxl  100010LL

char s1[maxl], s2[maxl];
int posl[maxl] = {0}, posr[maxl] = {0}, m, n, l, r, i, j, ansl, ansr;

int main() {
    scanf("%s%s", s1+1, s2+1);
    m = strlen(s1+1), n = strlen(s2+1);
    for (l = j = 1; l <= n; l ++ ) {
        for ( ; j <= m && !posl[l]; j ++ )
            if (s1[j] == s2[l]) posl[l] = j;
        if (!posl[l]) break;
    }
    for (r = n, j = m; r; r -- ) {
        for ( ; j && !posr[r]; j -- )
            if (s1[j] == s2[r]) posr[r] = j;
        if (!posr[r]) break;
    }
    if (l == 1 && r == n) puts("-");
    else if (l == 1) puts(s2+r+1);
    else if (r == n) s2[l] = '\0', puts(s2+1);
    else if (l == n+1) puts(s2+1);
    else {
        if (l-1 > n-r) ansl = l-1, ansr = n+1;
        else ansl = 0, ansr = r+1;
        j = r+1;
        for (i = 1; i < l; i ++ ) {
            while (j<=n && posr[j]<=posl[i]) j ++ ;
            if (j > n) break;
            if (i+n-j+1 > ansl+n-ansr+1)
                ansl = i, ansr = j;
        }
        for (i = 1; i <= ansl; i ++ ) putchar(s2[i]);
        for (i = ansr; i <= n; i ++ ) putchar(s2[i]);
            puts("");
    }
    getchar(),getchar();
    return 0;
}