CodeForcesGym 100548G The Problem to Slow Down You

The Problem to Slow Down You

Time Limit: 20000ms

Memory Limit: 524288KB

This problem will be judged on  CodeForcesGym. Original ID: 100548G
64-bit integer IO format: %I64d      Java class name: (Any)

 

After finishing his homework, our problem setter Federmann decided to kill time by hanging around online. He found a cool chat room that discusses competitive programming. Federmann has already joined lot of such chat rooms, but this one is special. Once he entered the  chat room, he noticed that there is an announcement saying “We forbid off-topic messages!”.  Federmann thinks that’s quite unusual, he decided to sit down and join the talk. After watching  people discussing different programming challenges for a while, he found an interesting  message saying “No, Federmann won’t prepare another problem about strings this year.”

“Oh, why do you guys think about that?” Federmann smiled. “Don’t they know I have an Edward number2 of 3?”
He then thought about something about palindrome, given two strings A and B, what is the number of their common palindrome substrings? The amount of common palindrome
substrings between two strings is defined as the number of quadruple (p, q, s, t), which satisfies that:

1. 1 ≤ p, q ≤ length(A), 1 ≤ s, t ≤ length(B), p ≤ q and s ≤ t. Here length(A) means the length of string A.

2. A_p..q = B_s..t

3. A_p..q is palindrome. (palindrome string is the string that reads the same forward or
backward) For example, (1, 3, 1, 3) and (1, 3, 3, 5) are both considered as a valid common palindrome
substring between aba and ababa. Federmann is excited about his new task, and he is just too lazy to write solutions, help
him.

Input
The first line of the input gives the number of test cases, T. T test cases follow. For each
test case, the first line contains a string A and the second line contains a string B. The length
of A, B will not exceed 200000.
It is guaranteed the input file will be smaller than 8 MB.

Output
For each test case, output one line containing “Case #x: y”, where x is the test case
number (starting from 1) and y is the number of common palindrome substrings of A and B.

Samples

Sample Input

3

abacab

abccab

faultydogeuniversity

hasnopalindromeatall

abbacabbaccab

youmayexpectedstrongsamplesbutnow

Sample Output

Case #1: 12

Case #2: 20

Case #3: 18

Source

2014-2015 ACM-ICPC, Asia Xian Regional Contest

解题：回文机

 1 #include 
 2 using namespace std;
 3 typedef long long LL;
 4 const int maxn = 200010;
 5 const int N = 26;
 6 struct PalindromicTree {
 7     int next[maxn][N],cnt[maxn],fail[maxn];
 8     int len[maxn],s[maxn],num[maxn];
 9     int tot,last,n;
10     int newnode(int length) {
11         memset(next[tot],0,sizeof next[tot]);
12         len[tot] = length;
13         cnt[tot] = num[tot] = 0;
14         return tot++;
15     }
16     int get_fail(int x) {
17         while(s[n - len[x] - 1] != s[n]) x = fail[x];
18         return x;
19     }
20     void init() {
21         last = tot = n = 0;
22         newnode(0);
23         newnode(-1);
24         fail[0] = 1;
25         s[n] = -1;
26     }
27     void count() {
28         for(int i = tot-1; i >= 0; --i)
29             cnt[fail[i]] += cnt[i];
30     }
31     void add(int c) {
32         c -= 'a';
33         s[++n] = c;
34         int cur = get_fail(last);
35         if(!next[cur][c]) {
36             int now = newnode(len[cur] + 2);
37             fail[now] = next[get_fail(fail[cur])][c];
38             next[cur][c] = now;
39             num[now] = num[fail[now]] + 1;
40         }
41         last = next[cur][c];
42         cnt[last]++;
43     }
44 } a,b;
45 LL ret;
46 void dfs(int u,int v) {
47     for(int i = 0; i < N; ++i) {
48         int x = a.next[u][i];
49         int y = b.next[v][i];
50         if(x && y) {
51             ret += (LL)a.cnt[x]*b.cnt[y];
52             dfs(x,y);
53         }
54     }
55 }
56 char str[maxn];
57 int main() {
58     int kase,cs = 1;
59     scanf("%d",&kase);
60     while(kase--) {
61         a.init();
62         b.init();
63         scanf("%s",str);
64         for(int i = 0; str[i]; ++i) a.add(str[i]);
65         scanf("%s",str);
66         for(int i = 0; str[i]; ++i) b.add(str[i]);
67         a.count();
68         b.count();
69         ret = 0;
70         dfs(0,0);
71         dfs(1,1);
72         printf("Case #%d: %I64d\n",cs++,ret);
73     }
74     return 0;
75 }

View Code

 

转载于:https://www.cnblogs.com/crackpotisback/p/4752773.html