题目链接

solution

用后缀数组来处理。将两个字符串拼接到一起，然后跑一边后缀数组。

后面的如果暴力做的话，就是枚举一下相同字符串的起始位置，然后用\(height\)数组求一下这两个后缀的\(LCP\)，答案加上\(LCP\)就行了。

然后考虑优化，枚举贡献，也就是枚举一个\(height_i\)找到以\(height_i\)为最小值的最长区间，该区间\(i\)两端分属两个字符串的位置都会两两产生贡献。找这个区间可以通过单调栈解决。然后就是统计贡献，只要分别于处理一下两个数组的前缀和就能统计这个区间内某个位置两端的分属两个字符串的对数。

code

/*
* @Author: wxyww
* @Date:   2020-04-20 18:55:57
* @Last Modified time: 2020-04-20 19:30:57
*/
#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 = 400010;
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 x[N],m,sa[N],height[N],y[N],c[N],n,bel[N];
char s[N],s1[N],s2[N];
void get_sa() {
	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);
		num = 0;
		x[sa[1]] = ++num;
		for(int i = 2;i <= n;++i) 
			x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? num : ++num;
		m = num;
		if(n == m) break;
	}
}
int rk[N];
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 sta[N],top,sum[3][N];
int get(int x,int l,int r) {
	return sum[x][r] - sum[x][max(0,l - 1)];
}
int main() {
	// freopen("1.in","r",stdin);
	scanf("%s",s1 + 1);
	scanf("%s",s2 + 1);
	for(int i = 1;s1[i];++i) {
		s[++n] = s1[i];
		bel[n] = 1;
	}

	s[++n] = '#';
	for(int i = 1;s2[i];++i) {
		s[++n] = s2[i];
		bel[n] = 2;
	}

	// printf("%s",s + 1);puts("");

	m = 'z';
	get_sa();

	// for(int i = 1;i <= n;++i) printf("%d ",sa[i]);puts("");

	get_height();

	// for(int i = 1;i <= n;++i) printf("%d ",height[i]);puts("");
	for(int i = 1;i <= n;++i) {
		sum[1][i] = sum[1][i - 1];
		sum[2][i] = sum[2][i - 1];
		sum[bel[sa[i]]][i]++;
	}
	ll ans = 0 ;
	for(int i = 1;i <= n + 1;++i) {
		while(top && height[sta[top]] > height[i]) {
			int j = sta[top - 1],k = sta[top];
			ans += 1ll * get(1,j,k - 1) * get(2,k,i - 1) * height[k];
			ans += 1ll * get(2,j,k - 1) * get(1,k,i - 1) * height[k];
			--top;
		}
		sta[++top] = i;
	}
	cout<<ans;


	return 0;
}

标签：get,int,top,height,HAOI2016,luogu3181,字符串,sa,include
来源： https://www.cnblogs.com/wxyww/p/luogu3181.html