Codeforces 213E Two Permutations 线段树 (看题解)

Two Permutations

关键是没想到按大小顺序把第二个排列一个一个加入线段树， 然后线段树维护整体的hash值， 

得到的hs值减去一个sub 之后与， 第一个排列的hash值比较。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

int n, m;
int a[N], b[N];
int pos[N];
ull Pow[N];

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

struct info {
    int cnt;
    ull hs;
};
info operator + (const info& a, const info& b) {
    info c;
    c.cnt = a.cnt + b.cnt;
    c.hs = b.hs + a.hs * Pow[b.cnt];
    return c;
}
struct segmentTree {
    info a[N << 2];
    void update(int p, int v, int l, int r, int rt) {
        if(l == r) {
            if(!v) a[rt].cnt = a[rt].hs = 0;
            else a[rt].cnt = 1, a[rt].hs = v;
            return;
        }
        int mid = l + r >> 1;
        if(p <= mid) update(p, v, lson);
        else update(p, v, rson);
        a[rt] = a[rt << 1] + a[rt << 1 | 1];
    }
} Tree;

int main() {
    for(int i = Pow[0] = 1; i < N; i++) Pow[i] = Pow[i - 1] * 233333;
    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", &b[i]);
        pos[b[i]] = i;
    }
    int ans = 0;
    ull hsA = 0;
    ull sub = 0;
    for(int i = 1; i <= n; i++) hsA *= 233333, hsA += a[i];
    for(int i = 1; i <= n; i++) sub *= 233333, sub += 1;
    for(int i = 1; i <= n; i++) Tree.update(pos[i], i, 1, m, 1);
    ans += hsA == Tree.a[1].hs;
    for(int i = n + 1; i <= m; i++) {
        Tree.update(pos[i - n], 0, 1, m, 1);
        Tree.update(pos[i], i, 1, m, 1);
        ull hsB = Tree.a[1].hs;
        hsB -= sub * (i - n);
        ans += hsA == hsB;
    }
    printf("%d
", ans);
    return 0;
 }

/*
*/