[UVA11019] Matrix Matcher AC自动机

本文主要是介绍[UVA11019] Matrix Matcher AC自动机，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

本来RK在这道题快得多的....

自动机就自动机吧

Pc[i]表示编号 Next[i]表示下一个相同行 Now在全局表示当前处理到了第几行

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
#define max(a, b) ((a) < (b) ? (b) : (a))
using namespace std;
typedef long long LL;
const int MAXN = 1000;
const int MAXL = 10000;
const int SZ = 26;
char s[MAXN+10][MAXN+10], P[MAXN/10+10][MAXN/10+10];
int ans, Now;
int Pc[MAXN+10], Next[MAXN+10], Len[MAXN+10];
int cnt[MAXN+10][MAXN+10];
struct Aho_Corasick
{int ch[MAXL+10][SZ], Ncnt, val[MAXL+10];int f[MAXL+10], last[MAXL+10];void init() { memset(ch[0], 0, sizeof(ch[0])); Ncnt = 1; }void Match(int pos, int v){int cur = Pc[v-1];int c = pos - Len[cur] + 1;while(cur >= 0) {if(Now >= cur)cnt[Now - cur][c]++;cur = Next[cur];}}void New() {memset(ch[Ncnt], 0, sizeof(ch[Ncnt]));val[Ncnt] = 0;}inline int ID(char c) { return c - 'a'; }void insert(char *s, int v){int u = 0, n = strlen(s);for(int i = 0; i < n; i++){int c = ID(s[i]);if(!ch[u][c]) {New();ch[u][c] = Ncnt++;}u = ch[u][c];}val[u] = v;}void getFail(){queue <int> q;f[0] = 0;for(int c = 0; c < SZ; c++){int u = ch[0][c];if(u) { f[u] = 0; q.push(u); last[u] = 0; }}while(!q.empty()){int r = q.front(); q.pop();for(int c = 0; c < SZ; 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]];}}}void add(int i, int j) {if(j) {Match(i, val[j]);add(i, last[j]);}}void solve(char *s) {int n = strlen(s), j = 0;for(int i = 0; i < n; i++){int c = ID(s[i]);j = ch[j][c];if(val[j]) add(i, j);else if(last[j]) add(i, last[j]);}}
}ac;
int main()
{int T; SF("%d", &T); while(T--) {int n, m, x, y;SF("%d%d", &n, &m);for(int i = 0; i < n; i++) SF("%s", s[i]);SF("%d%d", &x, &y);ac.init();for(int i = 0; i < x; i++) {SF("%s", P[i]);Len[i] = strlen(P[i]);Pc[i] = i;Next[i] = -1;for(int j = 0; j < i; j++) if(!strcmp(P[i], P[j])) {Pc[i] = j;Next[i] = Next[j];Next[j] = i;break;}if(Pc[i] == i) ac.insert(P[i], i + 1);}ac.getFail();memset(cnt, 0, sizeof(cnt));for(Now = 0; Now < n; Now++) ac.solve(s[Now]);ans = 0;for(int i = 0; i < n - x + 1; i++)for(int j = 0; j < m - y + 1; j++)if(cnt[i][j] == x) ans++;PF("%d\n", ans);}return 0;
}
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