【线段树+字符串hash】 codeforces 213E Two Permutations

对于排列单点插入到线段树中，然后判断整个线段树区间上的hash值和匹配串的hash值是否相同。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 200005
#define maxm 300005
#define eps 1e-10
#define mod 1000000007
#define INF 1e17
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

const int x = 123;
int a[maxn], b[maxn];
ULL sum[maxn<<2], xp[maxn], tt, t;
int cnt[maxn<<2];
int n, m, c, v, p;

void read(void)
{
	int xx;
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for(int i = 1; i <= m; i++) scanf("%d", &xx), b[xx] = i;
}

void build(int o, int L, int R)
{
	sum[o] = cnt[o] = 0;
	if(L == R) return;
	int mid = (L + R) >> 1;
	build(lson);
	build(rson);
}

void pushup(int o)
{
	cnt[o] = cnt[ls] + cnt[rs];
	sum[o] = sum[ls] + sum[rs] * xp[cnt[ls]];
}

void updata(int o, int L, int R)
{
	if(L == R) {
		cnt[o] += c, sum[o] += v;
		return;
	}
	int mid = (L + R) >> 1;
	if(p <= mid) updata(lson);
	else updata(rson);
	pushup(o);
}

void init(void)
{
	xp[0] = 1, tt = t = 0;
	for(int i = 1; i <= n; i++) xp[i] = xp[i-1] * x;
	for(int i = n; i > 0; i--) tt = tt * x + a[i];
	for(int i = 0; i < n; i++) t += xp[i];
	build(1, 1, m);
}

void work(void)
{
	int ans = 0;
	for(int i = 1; i <= m; i++) {
		if(i <= n) p = b[i], v = i, c = 1, updata(1, 1, m);
		else {
			p = b[i-n], v = -(i-n), c = -1, updata(1, 1, m);
			p = b[i], v = i, c = 1, updata(1, 1, m);
		}
		if(i >= n) {
			if(sum[1] == tt) ans++;
			tt += t;
		}
	}
	printf("%d\n", ans);
}

int main(void)
{
	read();
	init();
	work();
	return 0;
}