1031: [JSOI2007]字符加密Cipher
Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 7593 Solved: 3291
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Description
喜欢钻研问题的JS同学，最近又迷上了对加密方法的思考。一天，他突然想出了一种他认为是终极的加密办法
：把需要加密的信息排成一圈，显然，它们有很多种不同的读法。例如下图，可以读作：
JSOI07 SOI07J OI07JS I07JSO 07JSOI 7JSOI0把它们按照字符串的大小排序：07JSOI 7JSOI0 I07JSO JSOI07
OI07JS SOI07J读出最后一列字符：I0O7SJ，就是加密后的字符串（其实这个加密手段实在很容易破解，鉴于这是
突然想出来的，那就^^）。但是，如果想加密的字符串实在太长，你能写一个程序完成这个任务吗？
Input
输入文件包含一行，欲加密的字符串。注意字符串的内容不一定是字母、数字，也可以是符号等。
Output
输出一行，为加密后的字符串。
Sample Input
JSOI07
Sample Output
I0O7SJ
HINT
对于100%的数据字符串的长度不超过100000。
Source
析：用后缀数组存储的两次字符串，然后只要的按字典序输出答案就好。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e5 + 10;
const int maxm = 3e5 + 10;
const ULL mod = 3;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Array{
int s[maxn], sa[maxn], t[maxn], t2[maxn];
int h[maxn], r[maxn], c[maxn];
int n;
void init(){ n = 0; memset(sa, 0, sizeof sa); }
void build_sa(int m){
int *x = t, *y = t2;
for(int i = 0; i < m; ++i) c[i] = 0;
for(int i = 0; i < n; ++i) ++c[x[i] = s[i]];
for(int i = 1; i < m; ++i) c[i] += c[i-1];
for(int i = n-1; i >= 0; --i) sa[--c[x[i]]] = i;
for(int k = 1; k <= n; k <<= 1){
int p = 0;
for(int i = n-k; i < n; ++i) y[p++] = i;
for(int i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k;
for(int i = 0; i < m; ++i) c[i] = 0;
for(int i = 0; i < n; ++i) ++c[x[y[i]]];
for(int i = 1; i < m; ++i) c[i] += c[i-1];
for(int i = n-1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i];
swap(x, y);
p = 1; x[sa[0]] = 0;
for(int i = 1; i < n; ++i)
x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++;
if(p >= n) break;
m = p;
}
}
void getHight(){
int k = 0;
for(int i = 0; i < n; ++i) r[sa[i]] = i;
for(int i = 0; i < n; ++i){
if(k) --k;
int j = sa[r[i]-1];
while(s[i+k] == s[j+k]) ++k;
h[r[i]] = k;
}
}
};
char s[maxn];
Array arr;
int main(){
scanf("%s", s);
arr.init();
for(int i = 0; s[i]; ++i, ++n) arr.s[arr.n++] = s[i];
for(int i = 0; s[i]; ++i) arr.s[arr.n++] = s[i];
arr.s[arr.n++] = 0;
arr.build_sa(130);
arr.getHight();
for(int i = 1; i <= n*2; ++i)
if(arr.sa[i] < n) printf("%c", arr.s[arr.sa[i]+n-1]);
printf("
");
return 0;
}