JSOI 2012 玄武密码 (SAM)

一个比较套路的题目

因为从 \(SAM\) 的源点向下遍历 \(DAG\) 可以找到所有不相同的子串，所以我们直接跳就行了

#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
//#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define mp make_pair
#define pb push_back
#define pof pop_front
#define pob pop_back
#define fi first
#define se second
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(x".in", "r", stdin),freopen(x".out", "w", stdout)
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 20000010 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int MOD = 1000000007 ;
const double eps = 1e-7 ;
void douout(double x){ printf("%lf\n", x + 0.0000000001) ; }
template <class T> void print(T a) { cout << a << endl ; exit(0) ; }
template <class T> void chmin(T &a, T b) { if (a > b) a = b ; }
template <class T> void chmax(T &a, T b) { if (a < b) a = b ; }
template <class T> void upd(T &a, T b) { (a += b) %= MOD ; }
template <class T> void mul(T &a, T b) { a = (ll) a * b % MOD ; }

int n, Q, len, p, np, q, nq, last = 1, cnt = 1 ;
char s[N], t[N] ;

struct node {
	int len, fa, to[5] ;
} po[N] ;

void add(int c) {
	p = last ;
	np = last = ++cnt ;
	po[np].len = po[p].len + 1 ;
	for (; p && !po[p].to[c]; p = po[p].fa) po[p].to[c] = np ;
	if (!p) po[np].fa = 1 ;
	else {
		q = po[p].to[c] ;
		if (po[q].len == po[p].len + 1) po[np].fa = q ;
		else {
			nq = ++cnt ; po[nq] = po[q] ;
			po[nq].len = po[p].len + 1 ;
			po[np].fa = po[q].fa = nq ;
			for (; p && po[p].to[c] == q; p = po[p].fa) po[p].to[c] = nq ;
		}
	}
}

int f(char c) {
	if (c == 'W') return 0 ;
	if (c == 'E') return 1 ;
	if (c == 'N') return 2 ;
	if (c == 'S') return 3 ;
}

signed main(){
//	file("test") ;
	scanf("%d%d", &n, &Q) ;
    scanf("%s", s + 1) ;
    rep(i, 1, n) add(f(s[i])) ;
	while (Q--) {
		scanf("%s", t + 1) ; len = strlen(t + 1) ;
		int ans = 0, now = 1 ;
		rep(i, 1, len) {
			if (po[now].to[f(t[i])]) now = po[now].to[f(t[i])], ans++ ;
			else break ;
		}
		printf("%d\n", ans) ;
	}
	return 0 ;
}

/*
写代码时请注意：
	1.ll？数组大小，边界？数据范围？
	2.精度？
	3.特判？
	4.至少做一些
思考提醒：
	1.最大值最小->二分？
	2.可以贪心么？不行dp可以么
	3.可以优化么
	4.维护区间用什么数据结构？
	5.统计方案是用dp？模了么？
	6.逆向思维？
*/