Description

给以长度为 \(n\) 的字符串，要求每次只能从两边取一个字符，使得取出来之后字典序最小。

\(1\leq n\leq 30000\)

Solution

将字符串翻转后加在原字符串后，求个后缀排名。直接比较排名来判断取前还是取后。

Code

#include 
    
     using namespace std;const int N = (30000+5)<<1;char ch[N];int n, m, x[N<<1], y[N<<1], c[N], sa[N], rk[N];void get() {    for (int i = 1; i <= n; i++) c[x[i] = ch[i]]++;    for (int i = 2; i <= m; i++) c[i] += c[i-1];    for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;    for (int k = 1; k <= n; k <<= 1) {        int num = 0;        for (int i = n-k+1; i <= n; i++) y[++num] = i;        for (int i = 1; i <= n; i++) if (sa[i] > k) y[++num] = sa[i]-k;        for (int i = 1; i <= m; i++) c[i] = 0;        for (int i = 1; i <= n; i++) c[x[i]]++;        for (int i = 2; i <= m; i++) c[i] += c[i-1];        for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];        swap(x, y); x[sa[1]] = num = 1;        for (int i = 2; i <= n; i++)            x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? num : ++num;        if ((m = num) == n) break;    }    for (int i = 1; i <= n; i++) rk[sa[i]] = i;}void work() {    scanf("%d", &n); getchar();    for (int i = 1; i <= n; i++) {        scanf("%c", &ch[i]); getchar();        ch[(n<<1)+2-i] = ch[i];    }    ch[n+1] = '$'; n = (n<<1)+1; m = 100;    get();    for (int i = 1, lm = (n>>1), l = 1, r = lm; i <= lm; i++) {        if (rk[l] <= rk[n-r+1]) putchar(ch[l++]);        else putchar(ch[r--]);        if (i%80 == 0) puts("");    }}int main() {work(); return 0; }