Lost's revenge
64-bit integer IO format: %I64d Java class name: Main
One noon, when Lost was lying on the bed, the Spring Brother poster on the wall(Lost is a believer of Spring Brother) said hello to him! Spring Brother said, "I'm Spring Brother, and I saw AekdyCoin shames you again and again. I can't bear my believers were being bullied. Now, I give you a chance to rearrange your gene sequences to defeat AekdyCoin!".
It's soooo crazy and unbelievable to rearrange the gene sequences, but Lost has no choice. He knows some genes called "number theory gene" will affect one "level of number theory". And two of the same kind of gene in different position in the gene sequences will affect two "level of number theory", even though they overlap each other. There is nothing but revenge in his mind. So he needs you help to calculate the most "level of number theory" after rearrangement.
Input
For each testcase, first line is number of "number theory gene" N(1<=N<=50). N=0 denotes the end of the input file.
Next N lines means the "number theory gene", and the length of every "number theory gene" is no more than 10.
The last line is Lost's gene sequences, its length is also less or equal 40.
All genes and gene sequences are only contains capital letter ACGT.
Output
Sample Input
3 AC CG GT CGAT 1 AA AAA 0
Sample Output
Case 1: 3 Case 2: 2
Source
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 510; 4 struct Trie { 5 int ch[maxn][4],fail[maxn],cnt[maxn],tot; 6 int id[256]; 7 void init() { 8 tot = 0; 9 newnode(); 10 for(int i = 0; i < 4; ++i) id["ACGT"[i]] = i; 11 } 12 int newnode() { 13 memset(ch[tot],0,sizeof ch[tot]); 14 fail[tot] = cnt[tot] = 0; 15 return tot++; 16 } 17 void insert(char *str,int rt = 0) { 18 for(int i = 0; str[i]; ++i) { 19 int &x = ch[rt][id[str[i]]]; 20 if(!x) x = newnode(); 21 rt = x; 22 } 23 cnt[rt]++; 24 } 25 void build(int rt = 0) { 26 queue<int>q; 27 for(int i = 0; i < 4; ++i) 28 if(ch[rt][i]) q.push(ch[rt][i]); 29 while(!q.empty()) { 30 rt = q.front(); 31 q.pop(); 32 cnt[rt] += cnt[fail[rt]]; 33 for(int i = 0; i < 4; ++i) { 34 int &x = ch[rt][i],y = ch[fail[rt]][i]; 35 if(x) { 36 fail[x] = y; 37 q.push(x); 38 } else x = y; 39 } 40 } 41 } 42 int dp[maxn][11*11*11*11 + 5]; 43 int solve(char *str) { 44 int bt[4] = {0,0,0,1},num[4] = {}; 45 for(int i = 0; str[i]; ++i) ++num[id[str[i]]]; 46 for(int i = 2; i >= 0; --i) 47 bt[i] = bt[i + 1]*(num[i] + 1); 48 memset(dp,-1,sizeof dp); 49 int ans = dp[0][0] = 0; 50 for(int a = 0; a <= num[0]; ++a) 51 for(int b = 0; b <= num[1]; ++b) 52 for(int c = 0; c <= num[2]; ++c) 53 for(int d = 0; d <= num[3]; ++d) { 54 int hs = a*bt[0] + b*bt[1] + c*bt[2] + d; 55 for(int i = 0; i < tot; ++i) { 56 if(dp[i][hs] == -1) continue; 57 for(int k = 0; k < 4; ++k) { 58 if(a == num[0] && k == 0) continue; 59 if(b == num[1] && k == 1) continue; 60 if(c == num[2] && k == 2) continue; 61 if(d == num[3] && k == 3) continue; 62 int x = dp[ch[i][k]][hs + bt[k]]; 63 int y = dp[i][hs] + cnt[ch[i][k]]; 64 dp[ch[i][k]][hs + bt[k]] = max(x,y); 65 } 66 } 67 } 68 int st = num[0]*bt[0] + num[1]*bt[1] + num[2]*bt[2] + num[3]*bt[3]; 69 for(int i = 0; i < tot; ++i) 70 ans = max(ans,dp[i][st]); 71 return ans; 72 } 73 } ac; 74 char str[maxn]; 75 int main() { 76 int n,cs = 1; 77 while(scanf("%d",&n),n){ 78 ac.init(); 79 while(n--){ 80 scanf("%s",str); 81 ac.insert(str); 82 } 83 scanf("%s",str); 84 ac.build(); 85 printf("Case %d: %d\n",cs++,ac.solve(str)); 86 } 87 return 0; 88 }