HDU 3065 病毒侵袭持续中 AC自动机

思路：直接查找即可。

注意：如果不是’A’-‘Z’之间，直接赋值为26即可，因为反正也不会找到它的。

/*********************************************
Problem : HDU 2896
Author  : NMfloat
InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++) //遍历
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --) //反向遍历
#define repS(it,p) for(auto it = p.begin() ; it != p.end() ; it ++) //遍历一个STL容器
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next) //遍历u所连接的点
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m,k;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

const int MAXN = 50005;
const int MAX_SIZE = 27;
int tot ;
char s[2000005];
char sc[1005][55];
int virus[4];//需要输出病毒的编号
int recv[4];
int ans[1005];

class AC_automation {
private:
struct Node {
int next[MAX_SIZE]; //节点的儿子
int stat ; //有几个字符是以当前节点结尾的
}t[MAXN]; //Trie 树上的节点
int fail[MAXN]; //失配指针

public:
void init() {
tot = 1; cls(t,0); cls(fail,0);
}

void reset_stat(int len) {
rep(i,1,len) t[recv[i]].stat = virus[i];
}

void insert(char * str,int idx) {
int len = strlen(str) - 1;
rep(i,0,len) str[i] -= 'A';
int pre = 0 , nxt = 0 ;
rep(i,0,len) {
nxt = t[pre].next[str[i]];
if(nxt == 0) {
t[pre].next[str[i]] = tot ;
nxt = tot ++;
}
pre = nxt;
//printf("%d -> ",nxt);
}
//puts("");
t[nxt].stat = idx; //给每个字符串编号
rep(i,0,len) str[i] += 'A';
}

void get_fail() {
queue<int>q;
q.push(0);
while(!q.empty()) {
int u = q.front();
rep(i,0,MAX_SIZE-1) {
if(t[u].next[i]) {
q.push(t[u].next[i]);
int tmp = u;
while(fail[tmp]) {
if(t[fail[tmp]].next[i]) {
fail[t[u].next[i]] = t[fail[tmp]].next[i]; break;
}
tmp = fail[tmp];
}
if(u != 0 && !fail[t[u].next[i]]) {//处理fail[tmp] = 0 时的情况
if(t[0].next[i]) fail[t[u].next[i]] = t[0].next[i];
}
}
}
q.pop();
}
//rep(i,1,tot-1) printf("%d %d %d\n",i,fail[i],t[i].stat);
}

void query(char * str) {
int idx = 0;
int len = strlen(str) - 1;
rep(i,0,len) {
if(str[i] >= 'A' && str[i] <= 'Z') str[i] -= 'A';
else str[i] = 26;
}
int now = 0;
rep(i,0,len) {
while(now && (!t[now].next[str[i]])) now = fail[now];
now = t[now].next[str[i]];
int tmp = now;
while(tmp) {
if(t[tmp].stat) {
ans[t[tmp].stat] ++;
}
tmp = fail[tmp];
}
}
}
};

AC_automation ac;

void input() {
ac.init();
getchar();//读掉一个回车
rep(i,1,n) {
gets(sc[i]);
ac.insert(sc[i],i);
}
ac.get_fail();
}

void solve() {
scanf("%s",s);
cls(ans,0);
ac.query(s);
rep(i,1,n) {
if(ans[i]) {
printf("%s: %d\n",sc[i],ans[i]);
}
}
}

int main(void) {
//freopen("a.in","r",stdin);
while(~scanf("%d",&n)) {
input();
solve();
}
return 0;
}