【AC自动机+DP+高精度】Censored! _POJ1625
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Censored!
Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 8623 Accepted: 2341
Description
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 1abbb
Sample Output
5
Source
Northeastern Europe 2001, Northern Subregion
Kuang神说:
"AC自动机+DP+高精度
好题"
最近在啃AC自动机,代码理解上尚可,但是似乎有很多变型,多看多理解才是最重要的。
转载用意:DP学习+AC自动机(含矩阵加速)学习备看 & 大数据模板记录
Code By Kuangbin:
#include <queue>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>using namespace std;#include <map>map<char,int>mp;int N,M,P;struct Matrix{ int mat[110][110]; int n; Matrix(){} Matrix(int _n) { n=_n; for(int i = 0;i < n;i++) for(int j = 0;j < n;j++) mat[i][j] = 0; }};struct Trie{ int next[110][256], fail[110]; bool end[110]; int L,root; int newnode() { for(int i = 0;i < 256;i++) next[L][i] = -1; end[L++] = false; return L-1; } void init() { L = 0; root = newnode(); } void insert(char buf[]) { int len = strlen(buf); int now = root; for(int i = 0;i < len;i++) { if(next[now][mp[buf[i]]] == -1) next[now][mp[buf[i]]] = newnode(); now = next[now][mp[buf[i]]]; } end[now] = true; } void build() { queue<int>Q; fail[root] = root; for(int i = 0;i < 256;i++) if(next[root][i] == -1) next[root][i] = root; else { fail[next[root][i]] = root; Q.push(next[root][i]); } while(!Q.empty()) { int now = Q.front(); Q.pop(); if(end[fail[now]]==true)end[now]=true; for(int i = 0;i < 256;i++) if(next[now][i] == -1) next[now][i] = next[fail[now]][i]; else { fail[next[now][i]] = next[fail[now]][i]; Q.push(next[now][i]); } } } Matrix getMatrix() { Matrix res = Matrix(L); for(int i = 0;i < L;i++) for(int j = 0;j < N;j++) if(end[next[i][j]]==false) res.mat[i][next[i][j]]++; return res; } void debug() { for(int i = 0;i < L;i++) { printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]); for(int j = 0;j < 26;j++) printf("%2d",next[i][j]); printf("]\n"); } }};/* * 高精度,支持乘法和加法 */struct BigInt{ const static int mod = 10000; const static int DLEN = 4;//per 4 digit for a section int a[600],len; BigInt() { memset(a,0,sizeof(a)); len = 1; } BigInt(int v) { memset(a,0,sizeof(a)); len = 0; do { a[len++] = v%mod; v /= mod; }while(v); } BigInt(const char s[]) { memset(a,0,sizeof(a)); int L = strlen(s); len = L/DLEN; if(L%DLEN) len++; int index = 0; for(int i = L-1;i >= 0;i -= DLEN) { int t = 0,k = i - DLEN + 1; if(k < 0) k = 0; for(int j = k;j <= i;j++) t = t*10 + s[j] - '0'; a[index++] = t; } } BigInt operator +(const BigInt &b)const { BigInt res; res.len = max(len,b.len); for(int i = 0;i <= res.len;i++) res.a[i] = 0; for(int i = 0;i < res.len;i++) { res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0); res.a[i+1] += res.a[i]/mod; res.a[i] %= mod; } if(res.a[res.len] > 0)res.len++; return res; } BigInt operator *(const BigInt &b)const { BigInt res; for(int i = 0; i < len;i++) { int up = 0; for(int j = 0;j < b.len;j++) { int temp = a[i]*b.a[j] + res.a[i+j] + up; res.a[i+j] = temp%mod; up = temp/mod; } if(up != 0) res.a[i + b.len] = up; } res.len = len + b.len; while(res.a[res.len - 1] == 0 &&res.len > 1) res.len--; return res; } void output() { printf("%d",a[len-1]); for(int i = len-2;i >=0 ;i--) printf("%04d",a[i]); printf("\n"); }};char buf[1010];BigInt dp[2][110];Trie ac;int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout); while(scanf("%d%d%d",&N,&M,&P)==3) { gets(buf); gets(buf); mp.clear(); int len = strlen(buf); for(int i = 0;i < len;i++) mp[buf[i]]=i; ac.init(); for(int i = 0;i < P;i++) { gets(buf); ac.insert(buf); } ac.build();// ac.debug(); Matrix a= ac.getMatrix();// for(int i = 0;i <a.n;i++)// {// for(int j=0;j<a.n;j++)printf("%d ",a.mat[i][j]);// cout<<endl;// } int now = 0; dp[now][0] = 1; for(int i = 1;i < a.n;i++) dp[now][i] = 0; for(int i = 0;i < M;i++) { now^=1; for(int j = 0;j < a.n;j++) dp[now][j] = 0; for(int j = 0;j < a.n;j++) for(int k = 0;k < a.n;k++) if(a.mat[j][k] > 0) dp[now][k] = dp[now][k]+dp[now^1][j]*a.mat[j][k];// for(int j = 0;j < a.n;j++)// dp[now][j].output(); } BigInt ans = 0; for(int i = 0;i < a.n;i++) ans = ans + dp[now][i]; ans.output(); } return 0;}
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