luogu3763 [TJOI2017]DNA

题目链接

solution

直接暴力后缀数组。

将两个字符串接到一起，然后跑一遍后缀数组。

因为可以有3个字符不同，那我们就枚举第一个字符串中和第二个字符串第一个字符相匹配的位置。从这个位置开始每次跳\(LCP\)长度。看跳4次之后能不能跳到最后就行了。

求\(LCP\)长度的时候用\(ST\)表

code

/*
* @Author: wxyww
* @Date:   2020-04-20 20:00:41
* @Last Modified time: 2020-04-20 21:00:47
*/
#pragma GCC optimize ("O3")

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 200010,logN = 20;
ll read() {
	ll x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') f = -1; c = getchar();
	}
	while(c >= '0' && c <= '9') {
		x = x * 10 + c - '0'; c = getchar();
	}
	return x * f;
}
int n,m,sa[N],x[N],tn,y[N],c[N],st[N][logN + 2],rk[N],height[N];
char s[N];
void get_sa() {
	for(int i = 1;i <= m;++i) c[i] = 0;
	for(int i = 1;i <= n;++i) ++c[x[i] = s[i]];
	for(int i = 1;i <= m;++i) c[i] += c[i - 1];
	for(int i = n;i >= 1;--i) sa[c[x[i]]--] = i;
	for(int k = 1;k <= n;k <<= 1) {
		int num = 0;
		for(int i = n - k + 1;i <= n;++i) y[++num] = i;
		for(int i = 1;i <= n;++i) if(sa[i] > k) y[++num] = sa[i] - k;
		for(int i = 1;i <= m;++i) c[i] = 0;
		for(int i = 1;i <= n;++i) ++c[x[i]];
		for(int i = 1;i <= m;++i) c[i] += c[i - 1];
		for(int i = n;i >= 1;--i) sa[c[x[y[i]]]--] = y[i];
		swap(x,y);
		x[sa[1]] = num = 1;
		for(int i = 2;i <= n;++i)
			x[sa[i]] = (y[sa[i] + k] == y[sa[i - 1] + k] && y[sa[i]] == y[sa[i - 1]]) ? num : ++num;
		m = num;
		if(m == n) break;
	}
}
void get_height() {
	for(int i = 1;i <= n;++i) rk[sa[i]] = i;
	int k = 0;
	for(int i = 1;i <= n;++i) {
		if(rk[i] == 1) continue;
		if(k) --k;
		int j = sa[rk[i] - 1];
		while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) ++k;
		height[rk[i]] = k;
	}
}
int lg[N];
void get_st() {
	for(int i = 1;i <= n;++i) st[i][0] = height[i];
	for(int i = 2;i <= n;++i) lg[i] = lg[i >> 1] + 1;

	for(int j = 1;(1 << j) <= n;++j) {
		for(int i = 1;i <= n;++i) {
			st[i][j] = min(st[i][j - 1],st[i + (1 << (j - 1))][j - 1]);
		}
	}
}
int get(int l,int r) {
	if(l > r) swap(l,r);
	++l;
	return min(st[l][lg[r - l + 1]],st[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
}

void solve() {

	scanf("%s",s + 1);
	n = strlen(s + 1);
	tn = n;
	s[++n] = '#';
	scanf("%s",s + n + 1);
	n = strlen(s + 1);
	// printf("%s",s + 1);puts("");
	m = 'z';
	get_sa();
	get_height();
	get_st();
	// for(int i = 1;i <= n;++i) printf("%d ",sa[i]);puts("");
	// for(int i = 1;i <= n;++i) printf("%d ",height[i]);puts("");
	// puts("!!!");
	// cout<<get(6,10)<<endl;
	int ans = 0;
	for(int i = 1;i + n - tn - 2 <= tn;++i) {
		// printf("%d\n",i);
		int p = tn + 2;
		for(int j = 1;j <= 4 && p <= n;++j) {
			// if(i == 5) printf("!!%d %d %d\n",i + p  - tn - 2,p,get(rk[i + (p - tn - 2)],rk[p]));
			p += get(rk[i + (p - tn - 2)],rk[p]);
			if(j != 4) ++p;
		}
		if(p > n) {
			// cout<<i<<endl;
			ans++;
		}
	}
	printf("%d\n",ans);

}
int main() {
//	freopen("4892/7.in","r",stdin);
	int T = read();
	while(T--) {
		solve();
	}

	return 0;
}