题意：
给一个母串和多个模式串，求模式串在母串后翻转后的母串出现次数的的总和。
分析：
模板题
/*#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
const int maxnode = 250*1000+10000;
const int sigma_size = 26;
struct Trie{
int ch[maxnode][sigma_size];
int val[maxnode]; //该单词在模式串中出现的次数
int last[maxnode];
int f[maxnode]; //失配数组
int num[maxnode]; //该单词出现在文本串的次数
int pre[maxnode]; //该单词的前驱
int len[maxnode]; //以该单词结尾的单词长度
int Char[maxnode]; //该单词对应的字母
int road[maxnode]; //路径压缩优化 针对计算模式串出现的种数
int sz;
int Newnode()
{
val[sz] = f[sz] = last[sz] = len[sz] = num[sz] = 0;
memset(ch[sz], 0, sizeof ch[sz]);
return sz++;
}
void init(){
sz=0;
Newnode();
}
int idx(char c){ return c-‘A‘; }
int build(char *s){
int u = 0;
for(int i = 0, c; s[i] ;i++){
c = idx(s[i]);
if(!ch[u][c])
ch[u][c] = Newnode();
pre[ch[u][c]] = u;
Char[ch[u][c]] = s[i];
len[ch[u][c]] = len[u]+1;
road[ch[u][c]] = 1;
u = ch[u][c];
}
val[u] = 1;
num[u] = 0;
return u;
}
void getFail(){
queue<int> q;
for(int i = 0; i<sigma_size; i++)
if(ch[0][i]) q.push(ch[0][i]);
int r, c, u, v;
while(!q.empty()){
r = q.front(); q.pop();
for(c = 0; c<sigma_size; c++){
u = ch[r][c];
if(!u)continue;
q.push(u);
v = f[r];
while(v && ch[v][c] == 0) v = f[v]; //沿失配边走上去 如果失配后有节点 且 其子节点c存在则结束循环
f[u] = ch[v][c];
}
}
}
void find(char *T){
//计算模式串出现的个数：（每种多次出现算多次）
int j = 0;
for(int i = 0, c, temp; T[i] ; i++){
c = idx(T[i]);
while(j && ch[j][c]==0) j = f[j];
j = ch[j][c];
temp = j;
while(temp){
num[temp]++;
temp = f[temp];
}
}
}
int find_kind(char *T){
//计算种数， 重复出现的不再计算(若多个询问则要在此处加for(i=0->sz)lu[i]=1;
int j = 0, i, c, temp,ans=0;
for(i = 0; T[i]; i++){
c = idx(T[i]);
while(j && ch[j][c] == 0) j = f[j];
j = ch[j][c];
temp = j;
while(temp && road[temp]){
if(val[temp])
{
++ans;
val[temp] = 0;
}
road[temp] = 0;
temp = f[temp];
}
}
return ans;
}
}ac;
char s[1015], a[5100010], b[5100010], c;
int n, num;
int main() {
int T; scanf("%d", &T);
while (T-->0) {
ac.init();
scanf("%d", &n);
while(n--){
scanf("%s", s);
ac.build(s);
}
ac.getFail();
int top = 0;
scanf("%s", a);
int i = 0,sum=0,id=0;
while(a[i]!=‘\0‘){
if(a[i]==‘[‘){
i++;
int len=0;
while(a[i]>=‘0‘&&a[i]<=‘9‘){
len=len*10+a[i]-‘0‘;
i++;
}
for(int j=0;j<len;++j)
b[id++]=a[i];
i+=2;
}
else{
b[id++]=a[i];
i++;
}
}
b[id]=0;
//printf("%s\n",b);
sum+=ac.find_kind(b);
reverse(b,b+id);
//printf("%s\n",b);
sum+=ac.find_kind(b);
printf("%d\n", sum);
}
return 0;
}*/
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
#define N 300010
const ll INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod = 1000000007;
char P[1010],T[6000000],T1[6000000];
int n;
struct Trie{
int ch[N][26],val[N],f[N],num;
void init(){
num=1;
memset(ch,0,sizeof(ch));
memset(val,0,sizeof(val));
memset(f,0,sizeof(f));
}
void build(char *s){
int u=0,len=strlen(s);
for(int i=0;i<len;++i)
{
int v=s[i]-‘A‘;
if(!ch[u][v]){
memset(ch[num],0,sizeof(ch[num]));
ch[u][v]=num++;
}
u=ch[u][v];
}
val[u]=1;
//return u;
}
void getfail(){
queue<int>q;
for(int i=0;i<26;++i)
if(ch[0][i])
q.push(ch[0][i]);
while(!q.empty()){
int r=q.front();
q.pop();
for(int i=0;i<26;++i)
{
int u=ch[r][i];
if(!u){ch[r][i] = ch[f[r]][i];continue;}
q.push(u);
int v=f[r];
while(v&&!ch[v][i])v=f[v];
f[u]=ch[v][i];
}
}
}
int find(char *T){
int u=0,len=strlen(T),total=0;
for(int i=0;i<len;++i){
int v=T[i]-‘A‘;
while(u&&ch[u][v]==0)
u=f[u];
u=ch[u][v];
int tmp=u;
while(tmp){
if(val[tmp]){
total+=val[tmp];
val[tmp]=0;
}
tmp=f[tmp];
}
}
return total;
}
}ac;
int main()
{
int t;
scanf("%d",&t);
while(t--){
ac.init();
scanf("%d",&n);
while(n--){
scanf("%s",P);
ac.build(P);
}
scanf("%s",T);
// printf("%s",T);
int id=0,tid=0,i=0;
char tc;
while(T[i]!=‘\0‘){
if(T[i]==‘[‘){
i++;
int len=0;
while(T[i]>=‘0‘&&T[i]<=‘9‘){
len=len*10+T[i]-‘0‘;
i++;
}
for(int j=0;j<len;++j)
T1[id++]=T[i];
i+=2;
}
else{
T1[id++]=T[i];
i++;
}
}
T1[id]=0;
// printf("%s\n",T1);
int sum=0;
sum+=ac.find(T1);
reverse(T1,T1+id);
//printf("%s\n",T1);
sum+=ac.find(T1);
printf("%d\n",sum);
}
return 0;
}
时间： 2024-09-29 21:54:15