BZOJ 1031: [JSOI2007]字符加密Cipher 后缀数组

原题链接：

题解：

就把字符串扩大一倍后跑一发后缀数组即可。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#define INF 2*0x3f3f3f3f

using namespace std;
const int maxn = 200051; //注意数组的大小，记得更改

int wa[maxn], wb[maxn], wv[maxn], wss[maxn];
int cmp(int *r, int a, int b, int l) {
    return r[a] == r[b] && r[a + l] == r[b + l];
}
void da(int *r, int *sa, int n, int m) { //r为所求取的转化字符串 sa为求得的sa数组  n为数组长度+1  m为数组的最大值+1
    int i, j, p, *x = wa, *y = wb, *t;
    for (i = 0; i < m; i++) wss[i] = 0;
    for (i = 0; i < n; i++) wss[x[i] = r[i]]++;
    for (i = 1; i < m; i++) wss[i] += wss[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--wss[x[i]]] = i;
    for (j = 1, p = 1; p < n; j *= 2, m = p) {
        for (p = 0, i = n - j; i < n; i++) y[p++] = i;
        for (i = 0; i < n; i++) if (sa[i] >= j) y[p++] = sa[i] - j;
        for (i = 0; i < n; i++) wv[i] = x[y[i]];
        for (i = 0; i < m; i++) wss[i] = 0;
        for (i = 0; i < n; i++) wss[wv[i]]++;
        for (i = 1; i < m; i++) wss[i] += wss[i - 1];
        for (i = n - 1; i >= 0; i--) sa[--wss[wv[i]]] = y[i];
        for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
    return;
}

int ranking[maxn], height[maxn];
void build_height(int *r, int *sa, int n) { //r为求得的转化原始数组，sa为已经求得的数组，n为len
    int i, j, k = 0;
    for (i = 1; i <= n; i++) ranking[sa[i]] = i;
    for (i = 0; i < n; height[ranking[i++]] = k) {
        for (k ? k-- : 0, j = sa[ranking[i] - 1]; r[i + k] == r[j + k]; k++);
    }
}

int sa[maxn], r[maxn];

string s;
int str[maxn];

int main() {
    //freopen("1031.in","r",stdin);
    //freopen("1031.out","w",stdout);
    //cin.sync_with_stdio(false);
    cin >> s;
    int n = s.length();
    s = s + s;
    for (int i = 0; i < s.length(); i++)str[i] = s[i] + 2;
    str[s.length()]=1;
    da(str, sa, s.length() + 1, 230);
    for (int i = 0; i <= s.length(); i++) {
        if(sa[i]>=n)continue;
        int p = sa[i] + n - 1;
        cout << s[p];
    }
    cout << endl;
    return 0;
}