Palindromic Substring
Time Limit: 10000ms
Memory Limit: 65536KB
This problem will be judged on HDU. Original ID: 4426
64-bit integer IO format: %I64d Java class name: Main
In the kingdom of string, people like palindromic strings very much. They like only palindromic strings and dislike all other strings. There is a unified formula to calculate the score of a palindromic string. The score is calculated by applying the following three steps.
1. Since a palindromic string is symmetric, the second half (excluding the middle of the string if the length is odd) is got rid of, and only the rest is considered. For example, "abba" becomes "ab", "aba" becomes "ab" and "abacaba" becomes "abac".
2. Define some integer values for ‘a‘ to ‘z‘.
3. Treat the rest part as a 26-based number M and the score is M modulo 777,777,777.
However, different person may have different values for ‘a‘ to ‘z‘. For example, if ‘a‘ is defined as 3, ‘b‘ is defined as 1 and c is defined as 4, then the string "accbcca" has the score (3×263+4×262+4×26+1) modulo 777777777=55537.
One day, a very long string S is discovered and everyone in the kingdom wants to know that among all the palindromic substrings of S, what the one with the K-th smallest score is.
Input
The first line contains an integer T(1 ≤ T ≤ 20), the number of test cases.
The first line in each case contains two integers n, m (1 ≤ n ≤ 100000, 1 ≤ m ≤ 20) where n is the length of S and m is the number of people in the kingdom. The second line is the string S consisting of only lowercase letters. The next m lines each containing 27 integers describes a person in the following format.
Ki va vb ... vz
Where va is the value of ‘a‘ for the person, vb is the value of ‘b‘ and so on. It is ensured that the Ki-th smallest palindromic substring exists and va, vb, ..., vz are in the range of [0, 26). But the values may coincide.
Output
For each person, output the score of the K-th smallest palindromic substring in one line. Print a blank line after each case.
Sample Input
3 6 2 abcdca 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 7 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 4 10 zzzz 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 6 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 7 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 14 51 4 abcdefghijklmnopqrstuvwxyzyxwvutsrqponmlkjihgfedcba 1 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1 25 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1 26 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1 76 1 3 3 25 20 25 21 7 0 9 7 3 16 15 14 19 5 19 19 19 22 8 23 2 4 1
Sample Output
1 620 14 14 14 14 14 14 14 378 378 378 0 9 14 733665286
Hint
There are 7 palindromic substrings {"a", "a", "b", "c", "c", "d", "cdc"} in the first case. For the first person, the corresponding scores are {1, 1, 1, 1, 1, 1, 27}. For the second person, the corresponding scores are {25, 25, 24, 23, 23, 22, 620}.
Source
2012 Asia ChangChun Regional Contest
解题：麻辣隔壁，入坑了！k的范围会超过int，所以用long long较好！害我重写了两遍代码
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 500010;
4 const int mod = 777777777;
5 using PII = pair<int,int>;
6 using LL = long long;
7 int score[21][26];
8 PII d[maxn];
9 LL B[maxn],k[21];
10 struct PalindromicTree {
11 int ch[maxn][26],fail[maxn],cnt[maxn],len[maxn],s[maxn];
12 int tot,last,n,m;
13 LL hs[maxn][21];
14 void init() {
15 tot = last = n = 0;
16 newnode(0);
17 newnode(-1);
18 fail[0] = fail[1] = 1;
19 s[n] = -1;
20 }
21 int newnode(int slen = 0) {
22 memset(ch[tot],0,sizeof ch[tot]);
23 memset(hs[tot],0,sizeof hs[tot]);
24 fail[tot] = cnt[tot] = 0;
25 len[tot] = slen;
26 return tot++;
27 }
28 int getFail(int x) {
29 while(s[n - len[x] - 1] != s[n]) x = fail[x];
30 return x;
31 }
32 void extend(int c) {
33 s[++n] = c;
34 int cur = getFail(last);
35 if(!ch[cur][c]) {
36 int now = newnode(len[cur] + 2);
37 fail[now] = ch[getFail(fail[cur])][c];
38 ch[cur][c] = now;
39 int id = (len[cur] + 1)>>1;
40 for(int i = 0; i < m; ++i)
41 hs[now][i] = (hs[cur][i] + B[id]*score[i][s[n]])%mod;
42 }
43 ++cnt[last = ch[cur][c]];
44 }
45 void count() {
46 for(int i = tot-1; i > 1; --i)
47 cnt[fail[i]] += cnt[i];
48 }
49 int solve(int i){
50 LL tmp = 0;
51 int sz = 0;
52 for(int j = 2; j < tot; ++j)
53 d[sz++] = PII((int)hs[j][i],cnt[j]);
54 sort(d,d+sz);
55 for(int j = 0; j < sz; ++j){
56 tmp += d[j].second;
57 if(tmp >= k[i]) return d[j].first;
58 }
59 }
60 } pt;
61 char str[maxn];
62 int main() {
63 int kase,n,m;
64 for(int i = B[0] = 1; i < 100010; ++i) B[i] = B[i-1]*26%mod;
65 scanf("%d",&kase);
66 while(kase--) {
67 scanf("%d%d%s",&n,&m,str);
68 pt.init();
69 pt.m = m;
70 for(int i = 0; i < m; ++i) {
71 scanf("%I64d",k + i);
72 for(int j = 0; j < 26; ++j)
73 scanf("%d",score[i] + j);
74 }
75 for(int i = 0; i < n; ++i)
76 pt.extend(str[i]-‘a‘);
77 pt.count();
78 for(int i = 0; i < m; ++i)
79 printf("%d\n",pt.solve(i));
80 putchar(‘\n‘);
81 }
82 return 0;
83 }
84 /*
85 3
86 2 1
87 ab
88 1 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
89 */