标签:ace include algo substring comm front max cst ++i
题意:给定一个文本串和 n 个子串,问你子串在文本串出现的次数。
析:很明显的AC自动机,只要把先把子串进行失配处理,然后再去用文本串去匹配,在插入子串时就要标记每个串,注意串可能是相同的,这个我错了两次,最后匹配一次就OK了。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
char t[maxn];
char s[510][510];
const int maxnode = 510 * 510 + 10;
struct Aho{
int ch[maxnode][26];
int f[maxnode];
int val[maxnode];
int last[maxnode];
int cnt[510];
int sz;
void init(){
sz = 1;
memset(ch[0], 0, sizeof ch[0]);
memset(cnt, 0, sizeof cnt);
}
int idx(char c){ return c - ‘a‘; }
void insert(char *s, int v){
int u = 0;
while(*s){
int c = idx(*s);
if(!ch[u][c]){
memset(ch[sz], 0, sizeof ch[sz]);
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
++s;
}
val[u] = v;
}
void getFail(){
queue<int> q;
f[0] = 0;
for(int c = 0; c < 26; ++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 < 26; ++c){
int u = ch[r][c];
if(!u) 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 find(char *s){
int j = 0;
while(*s){
int c = idx(*s);
while(j && !ch[j][c]) j = f[j];
j = ch[j][c];
if(val[j]) print(j);
else if(last[j]) print(last[j]);
++s;
}
}
void print(int j){
if(j){
++cnt[val[j]];
print(last[j]);
}
}
};
Aho aho;
map<string, int> mp;
int main(){
int T; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d", &n);
scanf("%s", t);
aho.init();
mp.clear();
for(int i = 1; i <= n; ++i){
scanf("%s", s+i);
if(mp.count(s[i])) continue;
mp[s[i]] = i;
aho.insert(s[i], i);
}
aho.getFail();
aho.find(t);
printf("Case %d:\n", kase);
for(int i = 1; i <= n; ++i)
printf("%d\n", aho.cnt[mp[s[i]]]);
}
return 0;
}
LightOJ 1427 Substring Frequency (II) (AC自动机)
标签:ace include algo substring comm front max cst ++i
原文地址:http://www.cnblogs.com/dwtfukgv/p/7287208.html