HDU 482 String

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原文链接：http://www.cnblogs.com/crackpotisback/p/3938211.html

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#java

本文介绍了如何使用字符串哈希和动态映射来解决特定的字符串问题，包括如何识别和计算可恢复子串的数量。通过案例分析和详细步骤解释，深入探讨了算法的核心思想和实现过程。

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String

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 4821
64-bit integer IO format: %I64d Java class name: Main

Given a string S and two integers L and M, we consider a substring of S as “recoverable” if and only if
(i) It is of length M*L;
(ii) It can be constructed by concatenating M “diversified” substrings of S, where each of these substrings has length L; two strings are considered as “diversified” if they don’t have the same character for every position.

Two substrings of S are considered as “different” if they are cut from different part of S. For example, string "aa" has 3 different substrings "aa", "a" and "a".

Your task is to calculate the number of different “recoverable” substrings of S.

Input

The input contains multiple test cases, proceeding to the End of File.

The first line of each test case has two space-separated integers M and L.

The second ine of each test case has a string S, which consists of only lowercase letters.

The length of S is not larger than 10^5, and 1 ≤ M * L ≤ the length of S.

Output

For each test case, output the answer in a single line.

Sample Input

3 3
abcabcbcaabc

Sample Output

Source

2013 Asia Regional Changchun

解题：字符串hash+map

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #include <map>
14 #define LL long long
15 #define ULL unsigned long long
16 #define pii pair<int,int>
17 #define INF 0x3f3f3f3f
18 #define seek 131
19 using namespace std;
20 const int maxn = 100100;
21 map<ULL,int>mp;
22 char str[maxn];
23 ULL base[maxn],hs[maxn];
24 int main() {
25     int M,L,len,i,j,ans;
26     ULL tmp;
27     base[0] = 1;
28     for(i = 1; i < maxn; i++) base[i] = base[i-1]*seek;
29     while(~scanf("%d%d%s",&M,&L,str)){
30         len = strlen(str);
31         ans = 0;
32         hs[len] = 0;
33         for(i = len-1; i >= 0; i--)
34             hs[i] = hs[i+1]*seek+str[i]-'a';
35         for(i = 0; i < L && i + M*L <= len; i++){
36             mp.clear();
37             for(j = i; j < i+M*L; j += L){
38                 tmp = hs[j] - hs[j+L]*base[L];
39                 mp[tmp]++;
40             }
41             if(mp.size() == M) ans++;
42             for(j = i+M*L; j+L <= len; j += L){
43                 tmp = hs[j-M*L] - hs[j-M*L+L]*base[L];
44                 mp[tmp]--;
45                 if(!mp[tmp]) mp.erase(tmp);
46                 tmp = hs[j] - hs[j+L]*base[L];
47                 mp[tmp]++;
48                 if(mp.size() == M) ans++;
49             }
50         }
51         printf("%d\n",ans);
52     }
53     return 0;
54 }