BZOJ 1031: [JSOI2007]字符加密Cipher
2017-04-22 09:13
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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
JSOI07Sample Output
I0O7SJHINT
对于100%的数据字符串的长度不超过100000。分析
复习后缀数组的模板代码
#include <algorithm> #include <iostream> #include <cstring> #include <complex> #include <cstdio> #include <queue> #include <cmath> #include <map> #include <set> #define N 200005 #define INF 0x7fffffff #define sqr(x) ((x) * (x)) #define pi acos(-1) int len; int b ,c ,d ; int sa ,rank ,height ; char s , ans ; int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9'){if (ch == '-') f = -1; ch = getchar();} while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();} return x * f; } void getSa(int n,int m) { for (int i = 0; i <= m; i++) b[i] = 0; for (int i = 1; i <= n; i++) b[s[i]]++; for (int i = 1; i <= m; i++) b[i] += b[i - 1]; for (int i = n; i >= 1; i--) c[b[s[i]]--] = i; int t = 0; for (int i = 1; i <= n; i++) { if (s[c[i]] != s[c[i - 1]]) t++; rank[c[i]] = t; } int j = 1; while (j <= n) { for (int i = 0; i <= n; i++) b[i] = 0; for (int i = 1; i <= n; i++) b[rank[i + j]]++; for (int i = 1; i <= n; i++) b[i] += b[i - 1]; for (int i = n; i >= 1; i--) c[b[rank[i + j]]--] = i; for (int i = 0; i <= n; i++) b[i] = 0; for (int i = 1; i <= n; i++) b[rank[i]]++; for (int i = 1; i <= n; i++) b[i] += b[i - 1]; for (int i = n; i >= 1; i--) d[b[rank[c[i]]]--] = c[i]; t = 0; for (int i = 1; i <= n; i++) { if (rank[d[i]] != rank[d[i - 1]] || rank[d[i]] == rank[d[i - 1]] && rank[d[i - 1] + j] != rank[d[i] + j]) t++; c[d[i]] = t; } for (int i = 1; i <= n; i++) rank[i] = c[i]; if (t == n) break; j *= 2; } for (int i = 1; i <= n; i++) sa[rank[i]] = i; } void getHeight(int n) { int k = 0; for (int i = 1; i <= n; i++) { if (k) k--; int j = sa[rank[i] - 1]; while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) k++; height[rank[i]] = k; } } int main() { scanf("%s",s); int len = strlen(s), m = 0; for (int i = len; i >= 1; i--) { m = std::max(m,(int)s[i]); s[i + len] = s[i] = s[i - 1]; } getSa(len * 2, m); getHeight(len * 2); int j = 0; for (int i = 1; i <= len * 2; i++) if (sa[i] <= len) ans[++j] = s[sa[i] + len - 1]; for (int i = 1; i <= len; i++) printf("%c",ans[i]); printf("\n"); }
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