BZOJ 1031 [JSOI2007]字符加密Cipher (後綴數組)
阿新 • • 發佈:2017-11-11
問題 time rip air tput 數據 pan gre brush
Submit: 7593 Solved: 3291
[Submit][Status][Discuss]
1031: [JSOI2007]字符加密Cipher
Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 7593 Solved: 3291
[Submit][Status][Discuss]
Description
喜歡鉆研問題的JS同學,最近又迷上了對加密方法的思考。一天,他突然想出了一種他認為是終極的加密辦法 :把需要加密的信息排成一圈,顯然,它們有很多種不同的讀法。例如下圖,可以讀作:JSOI07 SOI07J OI07JS I07JSO 07JSOI 7JSOI0把它們按照字符串的大小排序:07JSOI 7JSOI0 I07JSO JSOI07 OI07JS SOI07J讀出最後一列字符:I0O7SJ,就是加密後的字符串(其實這個加密手段實在很容易破解,鑒於這是 突然想出來的,那就^^)。但是,如果想加密的字符串實在太長,你能寫一個程序完成這個任務嗎?
Input
輸入文件包含一行,欲加密的字符串。註意字符串的內容不一定是字母、數字,也可以是符號等。
Output
輸出一行,為加密後的字符串。
Sample Input
JSOI07Sample Output
I0O7SJHINT
對於100%的數據字符串的長度不超過100000。
Source
析:用後綴數組存儲的兩次字符串,然後只要的按字典序輸出答案就好。
代碼如下:
#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("\n"); return 0; }
BZOJ 1031 [JSOI2007]字符加密Cipher (後綴數組)