Bzoj4566：[HAOI2016]找相同字符

时间：2022-06-16 15:09:20

题面

Bzoj

Sol

两个串拼在一起后求出后缀数组

然后显然的\(n^2\)暴力，就是直接枚举求\(LCP\)

又由于扫的时候是对\(height\)取\(min\)

那么可以用单调栈维护每一段的贡献相同的

# include <bits/stdc++.h>

# define RG register

# define IL inline

# define Fill(a, b) memset(a, b, sizeof(a))

using namespace std;

typedef long long ll;

const int _(4e5 + 5);

int n, a[_], is[_], rk[_], sa[_], height[_], tmp[_], t[_];

int S[_], top, s1[_], s2[_];

ll ans, sum[_];

char ss[_];

IL int Cmp(RG int i, RG int j, RG int k){

	return tmp[i] == tmp[j] && tmp[i + k] == tmp[j + k] && i + k <= n && j + k <= n;

}

IL void Suffix_Sort(){

	RG int m = 27;

	for(RG int i = 1; i <= n; ++i) ++t[rk[i] = a[i]];

	for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

	for(RG int i = n; i; --i) sa[t[rk[i]]--] = i;

	for(RG int k = 1; k <= n; k <<= 1){

		RG int l = 0;

		for(RG int i = n - k + 1; i <= n; ++i) tmp[++l] = i;

		for(RG int i = 1; i <= n; ++i) if(sa[i] > k) tmp[++l] = sa[i] - k;

		for(RG int i = 0; i <= m; ++i) t[i] = 0;

		for(RG int i = 1; i <= n; ++i) ++t[rk[tmp[i]]];

		for(RG int i = 1; i <= m; ++i) t[i] += t[i - 1];

		for(RG int i = n; i; --i) sa[t[rk[tmp[i]]]--] = tmp[i];

		swap(rk, tmp), rk[sa[1]] = l = 1;

		for(RG int i = 2; i <= n; ++i) rk[sa[i]] = Cmp(sa[i - 1], sa[i], k) ? l : ++l;

		if(l >= n) break;

		m = l;

	}

	for(RG int i = 1, h = 0; i <= n; ++i){

		if(h) --h;

		while(a[i + h] == a[sa[rk[i] - 1] + h]) ++h;

		height[rk[i]] = h;

	}

}

int main(RG int argc, RG char* argv[]){

	scanf(" %s", ss);

	for(RG int i = 0, len = strlen(ss); i < len; ++i) a[++n] = ss[i] - 'a' + 1, is[n] = 1;

	scanf(" %s", ss), a[++n] = 27;

	for(RG int i = 0, len = strlen(ss); i < len; ++i) a[++n] = ss[i] - 'a' + 1, is[n] = 2;

	Suffix_Sort();

	for(RG int i = 1; i < n; ++i)

		s1[i] = s1[i - 1] + (is[sa[i]] == 1), s2[i] = s2[i - 1] + (is[sa[i]] == 2);

	S[0] = 1;

	for(RG int i = 1; i < n; ++i){

		while(top && height[S[top]] > height[i]) --top;

		S[++top] = i, sum[top] = sum[top - 1] + (s1[i - 1] - s1[S[top - 1] - 1]) * height[i];

		if(is[sa[i]] == 2) ans += sum[top];

	}

	top = 0;

	for(RG int i = 1; i < n; ++i){

		while(top && height[S[top]] > height[i]) --top;

		S[++top] = i, sum[top] = sum[top - 1] + (s2[i - 1] - s2[S[top - 1] - 1]) * height[i];

		if(is[sa[i]] == 1) ans += sum[top];

	}

	printf("%lld\n", ans);

    return 0;

}

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