Concatenation with intersection (2800 / z algorithm / tree array / double pointer)

Title: http://codeforces.com/contest/1313/problem/E
Reference: http://codeforces.com/blog/entry/74146
Title: The given string a, b, s. Find the number of substring combinations such that $a[l1,r1]+b[l2,r2]==s$ , request $[l1, r1], [l2, r2]$ intersection is not empty.
Solution: Use the z algorithm to find the longest prefix of a in s $lcp$ , the longest suffix of b in s $lcs$ . Enumerate a string from left to right, take $l_1$, Then the corresponding $r_2$The value is $l_1<=r_2<=l_1+m-2$ , we put all that satisfy the current interval $r_2$Corresponding leftmost endpoint $r_2-lcs_{r_2}$Throw them into a tree-like array, and count the number and sum respectively $cnt, I$ .
Then current $l_1$The number of cases for the left endpoint is $lcs*cnt-sum$， $sum$ part is not available and needs to be subtracted. This is only a roughly abstract description. There is a discrepancy between the code implementation and the theoretical description. See the code for details.

#include<bits/stdc++.h>
using namespace std;
const int maxn = 500010;
#define ll long long

int n,m;
char a[maxn],b[maxn];
char s[maxn*2],c[maxn*3];
int z[maxn*3];
int lcp[maxn];//lcp[i]表示a[i]开始,能匹配s的最长前缀 
int lcs[maxn];//lcs[i]表示b[i]开始,能匹配s的最长后缀 
//z算法，求解给定字符串每个位置能匹配自身的最长前缀 
void z_init(int len) {
	int l = 1,r = 1;
	z[1] = len;
	for(int i = 2;i <= len;i++) {
		if(i > r) {
			l = i;r = i;
			while(r<=len && c[r-i+1]==c[r]) r++;
			z[i] = r-l;r--;
		}else {
			int k = i-l+1;
			if(z[k]<r-i+1) z[i] = z[k];
			else {
				l = i;
				while(r<=len && c[r-i+1]==c[r]) r++;
				z[i] = r-l;r--;
			}
		}
	}
}
void init() {
	//求lcp 
	for(int i = 1;i <= m;i++) c[i] = s[i];
	c[m+1] = '#';
	for(int i = 1;i <= n;i++) c[m+1+i] = a[i];
	c[n+m+2] = '\0';
	z_init(n+m+1);
	for(int i = 1;i <= n;i++) lcp[i] = z[m+1+i];
	//求lcs
	for(int i = 1;i <= m;i++) c[i] = s[m+1-i];
	c[m+1] = '#';
	for(int i = 1;i <= n;i++) c[m+1+i] = b[n+1-i];
	c[n+m+2] = '\0';
	z_init(n+m+1);
	for(int i = 1;i <= n;i++) lcs[i] = z[m+1+n+1-i];
}
//Fenwick Tree
ll cnt[maxn*2],sum[maxn*2];
int lowbit(int x) {
	return x&(-x); 
} 
void add(int v) {
	int x = v;
	while(x <= n) sum[x]+=v,cnt[x]++,x+=lowbit(x);
}
void sub(int v) {
	int x = v;
	while(x <= n) sum[x]-=v,cnt[x]--,x+=lowbit(x);
}
ll get_sum(int x) {
	ll res = 0;
	while(x) res+=sum[x],x-=lowbit(x);
	return res;
}
ll get_cnt(int x) {
	ll res = 0;
	while(x) res+=cnt[x],x-=lowbit(x);
	return res;
} 
int main() {
	scanf("%d%d",&n,&m);
	scanf("%s%s%s",a+1,b+1,s+1);
	init();//puts("VE");
	ll ans = 0;
	/*
	*l1 <= r2 <= l1+m-2 as |r2-l1| <= m-1
	*initial l1 = 1
	*/
	for(int i = 1;i <= min(n,m-1);i++) add(max(1,m-lcs[i]));//puts("S");
	for(int i = 1,r;i <= n;i++) {
		r = min(m-1,lcp[i]);
		ans += 1LL*(r+1)*get_cnt(r)-get_sum(r);
		sub(max(1,m-lcs[i]));//delete case of "i as r2"
		if(i+m-1 <= n) add(max(1,m-lcs[i+m-1]));//add case of "i+m-1 as r2" 
	}
	printf("%I64d\n",ans); 
}