UVa11019 Matrix Matcher 留坑[AC自动机]

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val  int 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0  void 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fail    while(!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;}

来自：http://www.cnblogs.com/arbitrary/archive/2013/03/23/2977617.html
版权声明同资料整理系列。