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Problem H
Matrix Matcher
Input: Standard Input
Output: Standard Output
Given an N * M matrix, your task is to find the number of occurences of an X * Y pattern.
The first line contains a single integer t(t ≤ 15), the number of test cases.
For each case, the first line contains two integers N and M (N, M ≤ 1000). The next N lines contain M characters each.
The next line contains two integers X and Y (X, Y ≤ 100). The next X lines contain Y characters each.
For each case, output a single integer in its own line, the number of occurrences.
2 1 1 x 1 1 y 3 3 abc bcd cde 2 2 bc cd |
0 2
|
Problem Setter: Rujia Liu, EPS
Special Thanks: Wenbin Tang
Warming: The judge input file size is about 7 MB. So please make sure that you use a fast IO function (eg.scanf()) to read input.
AC自动机，把字符串P扔入AC自动机，然后去匹配T矩阵每行字符，
每次成功匹配，就把以当前行匹配算出来的矩阵的右上角cnt++，被+x次说明匹配矩阵成功。
注意vector中的元素用(*it)
PS：由于P中单词长度一致，故不用在print循环last
PS2:有可能同一行出现2次覆盖v，因此要用vector
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<queue>
#include<vector>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXT (15+10)
#define MAXN (1000+10)
#define MAXX (100+10)
#define MAXNode (1000000)
#define Sigma_size (26)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int cnt[MAXN][MAXN]; //右上角
class Aho_Corasick_Automata
{
public:
int ch[MAXNode][Sigma_size],siz;
vector<int> v[MAXNode];
// AC自动机
int f[MAXNode],last[MAXNode];
Aho_Corasick_Automata(int _siz=0):siz(_siz){MEM(ch) Rep(i,MAXNode) v[i].clear(); MEM(f) MEM(last)}
void mem(int _siz=0){siz=_siz; MEM(ch) Rep(i,MAXNode) v[i].clear(); MEM(f) MEM(last) }
int idx(char c){return c-'a';}
void insert(char *s,int val=1) //val!=0 表示单词末尾， PS:lrj叫它单词节点
{
int u=0,n=strlen(s);
Rep(i,n)
{
int c=idx(s[i]);
if (!ch[u][c])
{
++siz;
MEM(ch[siz]);
ch[u][c]=siz;
}
u=ch[u][c];
}
v[u].push_back(val);
}
void getFail()
{
queue<int> q;
Rep(c,Sigma_size)
{
int u=ch[0][c];
if (u) q.push(u),last[u]=0;
}
while (!q.empty())
{
int r=q.front();q.pop(); //r--c-->u
Rep(c,Sigma_size)
{
int u=ch[r][c];
if (!u) {ch[r][c]=ch[f[r]][c]; continue;}
q.push(u);
f[u]=ch[f[r]][c];
last[u]=v[f[u]].size()?f[u]:last[f[u]];
}
}
}
void print(int j,int r,int c) //打印全串中所有以j为末尾的str
{
for(vector<int>::iterator it=v[j].begin();it!=v[j].end();it++)
{
int P_i=(*it);
if (r-(P_i-1)<0) continue ;
cnt[r-(P_i-1)][c]++;
}
}
void find(char *s,int r)
{
int u=0,n=strlen(s);
Rep(i,n)
{
int c=idx(s[i]);
u=ch[u][c];
if (v[u].size()) print(u,r,i);
else if (last[u]) print(u,r,i);
}
}
}T;
int n,m,x,y;
char s[MAXN][MAXN];
char s2[MAXN];
int main()
{
// freopen("uva11019.in","r",stdin);
// freopen(".out","w",stdout);
int tt;
scanf("%d",&tt);
while(tt--)
{
T.mem();
scanf("%d%d",&n,&m);
Rep(i,n)
{
scanf("%s",s[i]);
}
scanf("%d%d",&x,&y);
For(i,x)
{
scanf("%s",s2);
T.insert(s2,i);
}
T.getFail();
MEM(cnt)
Rep(i,n)
{
char *str=s[i];
T.find(str,i);
}
int ans=0;
Rep(i,n)
{
Rep(j,m)
ans+=(bool)(cnt[i][j]==x);
}
cout<<ans<<endl;
}
return 0;
}
UVA 11019(Matrix Matcher-vector从迭代器中取值，AC自动机匹配字符矩阵)
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原文地址：http://blog.csdn.net/nike0good/article/details/44161285
