codeforces 633C. Spy Syndrome 2 hash
2016-03-07 12:46
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题目链接
C. Spy Syndrome 2
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined a new cipher technique.
For a given sentence, the cipher is processed as:
Convert all letters of the sentence to lowercase.
Reverse each of the words of the sentence individually.
Remove all the spaces in the sentence.
For example, when this cipher is applied to the sentence
Kira is childish and he hates losing
the resulting string is
ariksihsidlihcdnaehsetahgnisol
Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 10 000) — the length of the ciphered text. The second line consists of nlowercase English letters — the ciphered text t.
The third line contains a single integer m (1 ≤ m ≤ 100 000) — the number of words which will be considered while deciphering the text. Each of the next m lines contains a non-empty word wi (|wi| ≤ 1 000) consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1 000 000.
Output
Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those.
Examples
input
output
input
output
C. Spy Syndrome 2
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
After observing the results of Spy Syndrome, Yash realised the errors of his ways. He now believes that a super spy such as Siddhant can't use a cipher as basic and ancient as Caesar cipher. After many weeks of observation of Siddhant’s sentences, Yash determined a new cipher technique.
For a given sentence, the cipher is processed as:
Convert all letters of the sentence to lowercase.
Reverse each of the words of the sentence individually.
Remove all the spaces in the sentence.
For example, when this cipher is applied to the sentence
Kira is childish and he hates losing
the resulting string is
ariksihsidlihcdnaehsetahgnisol
Now Yash is given some ciphered string and a list of words. Help him to find out any original sentence composed using only words from the list. Note, that any of the given words could be used in the sentence multiple times.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 10 000) — the length of the ciphered text. The second line consists of nlowercase English letters — the ciphered text t.
The third line contains a single integer m (1 ≤ m ≤ 100 000) — the number of words which will be considered while deciphering the text. Each of the next m lines contains a non-empty word wi (|wi| ≤ 1 000) consisting of uppercase and lowercase English letters only. It's guaranteed that the total length of all words doesn't exceed 1 000 000.
Output
Print one line — the original sentence. It is guaranteed that at least one solution exists. If there are multiple solutions, you may output any of those.
Examples
input
30 ariksihsidlihcdnaehsetahgnisol 10 Kira hates is he losing death childish L and Note
output
Kira is childish and he hates losing
input
12 iherehtolleh 5 HI Ho there HeLLo hello
output
HI there HeLLo 把每一个小的串hash一下, 然后就可以在大的串里面找了, 具体看代码, 很清楚。
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> #include <stack> #include <bitset> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, n, a) for(int i = a; i<n; i++) #define fi first #define se second typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; map <ll, string> mp; ll dp[10005]; void print(int n) { if(!n) return ; string tmp = mp[dp ]; print(n-tmp.size()); cout<<tmp<<" "; } int main() { int n, m; string s, str; cin>>n>>s>>m; for(int i = 0; i<m; i++) { cin>>str; int len = str.size(); ll tmp = 0; for(int j = 0; j<len; j++) { tmp = tmp*31+tolower(str[j]); } mp[tmp] = str; } mem1(dp); dp[0] = 0; int len = s.size(); for(int i = 0; i<len; i++) { ll tmp = 0; for(int j = i; j>=0; j--) { tmp = tmp*31+s[j]; if(dp[j]!=-1 && mp.count(tmp)) { dp[i+1] = tmp; break; } } } int i = 0; print(len); return 0; }
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