poj 3623(后缀数组)

poj 3623
字符串翻转后的和原串后的数组进行求后缀数组，然后之后两个指针$i，j$选择两端$rk$值最小的输出，注意格式
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
#define ll long long
using namespace std;

const int MAXN = 30000 * 2 + 5; 

int N, n, m, a[MAXN];
int SA[MAXN], rk[MAXN], tp[MAXN], tax[MAXN];

bool cmp(int *f, int i, int k) {return f[SA[i]]==f[SA[i-1]]&&f[SA[i]+k]==f[SA[i-1]+k];}
void build() {
    for (int i=0;i<m;i++) tax[i] = 0;
    for (int i=0;i<n;i++) tax[rk[i]=a[i]]++;
    for (int i=1;i<m;i++) tax[i] += tax[i-1];
    for (int i=n-1;i>=0;i--) SA[--tax[rk[i]]] = i;
    int p;
    for (int k=1;k<=n;k*=2) {
        p = 0;
        for (int i=n-k;i<n;i++) tp[p++] = i;
        for (int i=0;i<n;i++) if (SA[i]>=k) tp[p++] = SA[i] - k;
        for (int i=0;i<m;i++) tax[i] = 0;
        for (int i=0;i<n;i++) tax[rk[tp[i]]]++;
        for (int i=1;i<m;i++) tax[i] += tax[i-1];
        for (int i=n-1;i>=0;i--) SA[--tax[rk[tp[i]]]] = tp[i];
        swap(tp, rk), p = 0, rk[SA[0]] = 0;
        for (int i=1;i<n;i++) rk[SA[i]] = cmp(tp, i, k) ? p : ++p;
        if (++p>=n) break;
        m = p;
    }
}
void init() {
    char ch[10];
    n = N;
    for (int i=0;i<N;i++) {
        scanf("%s", ch);
        a[i] = ch[0];
    }    
    a[n++] = '$';
    for (int i=N-1;i>=0;i--) {
        a[n++] = a[i];
    }    
    a[n++] = 0;
    m = 128;
}
void solve() {
    build();
    int i = 0, j = N+1, tot = 0;
    do {
        if (rk[i]<rk[j]) putchar(a[i]), i++, tot++; 
            else putchar(a[j]), j++, tot++;
            if (tot%80==0) printf("\n");
    } while (tot<N);
    printf("\n");
}
int main() {
    #ifndef ONLINE_JUDGE
    freopen("1.in", "r", stdin);freopen("1.out", "w", stdout);
    #endif
    while (scanf("%d", &N)==1&&N) init(), solve();
    return 0;
}