【BZOJ 1031】[JSOI2007]字符加密Cipher(后缀数组模板)

【题目链接】:http://www.lydsy.com/JudgeOnline/problem.php?id=1031

【题意】

【题解】

后缀数组模板题;
把整个字符串扩大一倍.
即长度乘2
然后搞出后缀数组;
然后顺序枚举i;
对于sa[i]< n的输出对应的s[sa[i]+n-1]就好了
后缀的含义是把后缀按照字典序从小到大排一下.
按照这个规则;
就能搞了;
必然是在前n个字符中就能比较出来;
所以及时后缀多了一些也没事.

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 1e5+100;

char s[N * 2];
int sa[N * 2], wv[2][350005], tong[N * 2];

bool cmp(int *tmp, int x, int y, int j)
{
    return tmp[x] == tmp[y] && tmp[x + j] == tmp[y + j];
}

void getsa(int n, int m)
{
    int p = 0, *x = wv[0], *y = wv[1];
    rep1(i, 0, n - 1)
        ++tong[x[i] = s[i]];
    rep1(i, 1, m - 1)
        tong[i] += tong[i - 1];
    rep2(i, n - 1, 0)
        sa[--tong[x[i]]] = i;
    for (int j = 1; p != n; j <<= 1, m = p)
    {
        p = 0;
        rep1(i,n-j,n-1)
            y[p++] = i;
        rep1(i, 0, n - 1)
            if (sa[i] >= j)
                y[p++] = sa[i] - j;
        rep1(i, 0, m - 1)
            tong[i] = 0;
        rep1(i, 0, n - 1)
            ++tong[x[y[i]]];
        rep1(i, 1, m - 1)
            tong[i] += tong[i - 1];
        rep2(i, n - 1, 0)
            sa[--tong[x[y[i]]]] = y[i];
        swap(x, y), p = 1, x[sa[0]] = 0;
        rep1(i, 1, n - 1)
            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;
    }
}

int main()
{
    //freopen("F:\\rush.txt", "r", stdin);
    int n;
    scanf("%s", s);
    n = strlen(s);
    rep1(i, 0, n - 1)
        s[i + n] = s[i];
    getsa(n << 1 | 1, 128);
    rep1(i, 1, n << 1)
        if (sa[i] < n)
            putchar(s[sa[i] + n - 1]);
    puts("");
    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}
posted @ 2017-10-04 18:45  AWCXV  阅读(117)  评论(0编辑  收藏  举报