UVa11019 Matrix Matcher 留坑[AC自动机]

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声明：此题没写，只理解了一下思路，坑留在这里什么时候复习再做。
Given an N M matrix, your task is to nd the number of occurences of an X Y pattern.
Input
The rst line contains a single integer t (t 15), the number of test cases.
For each case, the rst 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.
Output
For each case, output a single integer in its own line, the number of occurrences.
Sample Input
2
1 1
x
1 1
y
3 3
abc
bcd
cde
2 2
bc
cd
Sample Output
0
2

题意：给定二维文本串二维模板串，求匹配数；
分析：二维文本串比较短…所以…其实就是对模板串枚举行逐行匹配，用文本串建AC自动机，这样对于每一行都能知道在哪一行出现过（AC自动机并没有直接把两个单词拆分组合，这一点要注意，每个走出来的单词都是完整的），然后f记录一下再枚举相加就可以了。
标程：

// File Name: 11019.cpp
// Author: zlbing
// Created Time: 2013/3/23 14:37:07#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int SIGMA_SIZE = 26;
const int MAXNODE = 111000;
const int MAXS = 150 + 10;map<string,int> ms;
//ms是为了满足特殊要求,比如模板串相同时
struct ACautomata {int ch[MAXNODE][SIGMA_SIZE];int f[MAXNODE];    // fail函数int val[MAXNODE];  // 每个字符串的结尾结点都有一个非0的valint last[MAXNODE]; // 输出链表的下一个结点int next[MAXS];int sz;int d[1005][1005];void init() {sz = 1;memset(ch[0], 0, sizeof(ch[0]));ms.clear();memset(d,0,sizeof(d));memset(next,0,sizeof(next));}// 字符c的编号int idx(char c) {return c-'a';}// 插入字符串。v必须非0void insert(char *s, int v) {int u = 0, n = strlen(s);for(int i = 0; i < n; i++) {int c = idx(s[i]);if(!ch[u][c]) {memset(ch[sz], 0, sizeof(ch[sz]));val[sz] = 0;ch[u][c] = sz++;}u = ch[u][c];}if(val[u]){next[v]=val[u];}val[u] = v;ms[string(s)] = v;}// 递归打印匹配文本串str[i]结尾的后缀,以结点j结尾的所有字符串void print(int i,int j,int x) {if(j) {if(x-val[j]+1>0)d[x-val[j]+1][i]++;int t=val[j];while(next[t]){t=next[t];if(x-t+1>0)d[x-t+1][i]++;}print(i,last[j],x);}}// 在T中找模板int find(char* T,int x) {int n = strlen(T);int j = 0; // 当前结点编号，初始为根结点for(int i = 0; i < n; i++) { // 文本串当前指针int c = idx(T[i]);j = ch[j][c];if(val[j]) print(i,j,x);else if(last[j]) print(i,last[j],x); // 找到了！}}// 计算fail函数void getFail() {queue<int> q;f[0] = 0;// 初始化队列for(int c = 0; c < SIGMA_SIZE; c++) {int u = ch[0][c];if(u) { f[u] = 0; q.push(u); last[u] = 0; }}// 按BFS顺序计算failwhile(!q.empty()) {int r = q.front(); q.pop();for(int c = 0; c < SIGMA_SIZE; c++) {int u = ch[r][c];if(!u) {ch[r][c]=ch[f[r]][c];continue;}q.push(u);int v = f[r];while(v && !ch[v][c]) v = f[v];f[u] = ch[v][c];last[u] = val[f[u]] ? f[u] : last[f[u]];}}}};
ACautomata solver;
char str[1005][1005];
char str1[105][105];
int main(){int T;scanf("%d",&T);while(T--){int N,M,X,Y;scanf("%d%d",&N,&M);REP(i,1,N)scanf("%s",str[i]);scanf("%d%d",&X,&Y);REP(i,1,X)scanf("%s",str1[i]);solver.init();REP(i,1,X){solver.insert(str1[i],i);}solver.getFail();REP(i,1,N){solver.find(str[i],i);}int ans=0;REP(i,1,N)REP(j,1,M)if(solver.d[i][j]==X)ans++;printf("%d\n",ans);}return 0;
}