Censored!
Time Limit: 5000ms
Memory Limit: 10000KB
This problem will be judged on PKU. Original ID: 162564-bit integer IO format: %lld Java class name: Main
The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.
Input
The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).
The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.
The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.
Output
Output the only integer number -- the number of different sentences freelanders can safely use.
Sample Input
2 3 1 ab bb
Sample Output
5
Source
解题:Trie图 + 动态规划
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 struct BigInt { 7 const static int mod=10000; 8 const static int dlen=4; 9 int a[110],len; 10 11 BigInt() { 12 memset(a,0,sizeof(a)); 13 len=1; 14 } 15 16 BigInt(int v) { 17 memset(a,0,sizeof(a)); 18 len=0; 19 do { 20 a[len++]=v%mod; 21 v/=mod; 22 } while(v); 23 } 24 25 BigInt operator + (const BigInt &b) const { 26 BigInt res; 27 res.len=max(len,b.len); 28 for(int i=0; i<=res.len; i++) { 29 res.a[i]=0; 30 } 31 for(int i=0; i<res.len; i++) { 32 res.a[i]+=((i<len)?a[i]:0)+((i<b.len)?b.a[i]:0); 33 res.a[i+1]+=res.a[i]/mod; 34 res.a[i]%=mod; 35 } 36 if(res.a[res.len]>0) res.len++; 37 return res; 38 } 39 40 void output() { 41 printf("%d",a[len-1]); 42 for(int i=len-2; i>=0; i--) 43 printf("%04d",a[i]); 44 printf(" "); 45 } 46 } dp[110][110]; 47 const int maxn = 256; 48 struct Trie{ 49 int ch[maxn][maxn],fail[maxn],cnt[maxn],tot,n; 50 int mp[256]; 51 int newnode(){ 52 memset(ch[tot],0,sizeof ch[tot]); 53 fail[tot] = cnt[tot] = 0; 54 return tot++; 55 } 56 void init(char *str,int n){ 57 tot = 0; 58 for(int i = 0; i < n ; ++i) mp[str[i]] = i; 59 this->n = n; 60 newnode(); 61 } 62 void insert(char *str,int root = 0){ 63 for(int i = 0; str[i]; ++i){ 64 int &x = ch[root][mp[str[i]]]; 65 if(!x) x = newnode(); 66 root = ch[root][mp[str[i]]]; 67 } 68 ++cnt[root]; 69 } 70 void build(int root = 0){ 71 queue<int>q; 72 for(int i = 0; i < n; ++i) 73 if(ch[root][i]) q.push(ch[root][i]); 74 while(!q.empty()){ 75 root = q.front(); 76 q.pop(); 77 cnt[root] += cnt[fail[root]]; 78 for(int i = 0; i < n; ++i){ 79 int &x = ch[root][i],y = ch[fail[root]][i]; 80 if(x){ 81 fail[x] = y; 82 q.push(x); 83 }else x = y; 84 } 85 } 86 } 87 void solve(int m){ 88 for(int i = 0; i < 110; ++i) 89 for(int j = 0; j < 110; ++j) 90 dp[i][j] = BigInt(0); 91 dp[0][0] = BigInt(1); 92 for(int i = 1; i <= m; ++i){ 93 for(int j = 0;j < tot; ++j){ 94 if(cnt[j]) continue; 95 for(int k = 0; k < n; ++k){ 96 int x = ch[j][k]; 97 if(cnt[x]) continue; 98 dp[i][x] = dp[i][x] + dp[i-1][j]; 99 } 100 } 101 } 102 BigInt ans = BigInt(0); 103 for(int i = 0; i < tot; ++i) 104 ans = ans + dp[m][i]; 105 ans.output(); 106 } 107 }ac; 108 int main(){ 109 int n,m,p; 110 char str[maxn]; 111 while(~scanf("%d%d%d",&n,&m,&p)){ 112 scanf("%s",str); 113 ac.init(str,n); 114 while(p--){ 115 scanf("%s",str); 116 ac.insert(str); 117 } 118 ac.build(); 119 ac.solve(m); 120 } 121 return 0; 122 }