BZOJ4358: permu(带撤销并查集 不删除莫队)

题意

题目链接

sol

感觉自己已经老的爬不动了。。

想了一会儿，大概用个不删除莫队+带撤销并查集就能搞了吧，\(n \sqrt{n} logn\)应该卡的过去

不过不删除莫队咋写来着？。。。。跑去学。。

带撤销并查集咋写来着？。。。。跑去学。。。

发现自己的带撤销并查集是错的，，自己yy着调了1h终于过了大数据。。

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
#define pb(x) push_back(x)
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    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 n, m, a[maxn], belong[maxn], block, ans[maxn], cnt, fa[maxn];
struct q {
    int l, r, id;
    bool operator < (const q &rhs) const{
        return r < rhs.r;
    }
};
vector<q> q[maxn];
int solveblock(int x, int y) {
    if(x == y) return 1;
    vector<int> v;
    for(int i = x; i <= y; i++) v.pb(a[i]);
    sort(v.begin(), v.end());
    int res = 1, now = 1; 
    for(int i = 1; i < v.size(); i++) 
        now = (v[i] == v[i - 1] + 1 ? now + 1 : 1), chmax(res, now);
    return res;
}
int inder[maxn], top, ha[maxn], cur, mx;
struct node {
    int x, deg;
}s[maxn];
int find(int x) {
    return fa[x] == x ? x : find(fa[x]);
}
void unionn(int x, int y) {
    x = find(x); y = find(y);
    if(x == y) return;
    if(inder[x] < inder[y]) swap(x, y);
    chmax(mx, inder[x] + inder[y]);
    fa[y] = x;
    s[++top] = (node) {y, inder[y]};
    s[++top] = (node) {x, inder[x]};//tag
    inder[x] += inder[y];
}
void delet(int cur) {
    while(top > cur) {
        node pre = s[top--];
        fa[pre.x] = pre.x;
        inder[pre.x] = pre.deg;
    }
}
void add(int x) {
    ha[x] = 1;
    if(ha[x - 1]) unionn(x - 1, x);
    if(ha[x + 1]) unionn(x, x + 1);
}
void solve(int i, vector<q> &v) {
    memset(ha, 0, sizeof(ha));
    top = 0; int r = min(n, i * block) + 1;
    int ql = r, qr = ql - 1;//tag
    cur = 0, mx = 1;
    for(int i = 1; i <= n; i++) fa[i] = i, inder[i] = 1;
    for(int i = 0; i < v.size(); i++) {
        q x = v[i];
        while(qr < x.r) add(a[++qr]);
        cur = mx; int pre = top;
        while(ql > x.l) add(a[--ql]);
        ans[x.id] = mx;
        mx = cur;
        delet(pre);
        while(ql < r) ha[a[ql++]] = 0;
    }
}
signed main() {
    int mx = 0;
    n = read(); m = read(); block = sqrt(n); 
    for(int i = 1; i <= n; i++) a[i] = read(), belong[i] = (i - 1) / block + 1, chmax(mx, belong[i]);
    for(int i = 1; i <= m; i++) {
        int x = read(), y = read();
        if(belong[x] == belong[y]) ans[i] = solveblock(x, y);
        else q[belong[x]].push_back({x, y, i});
    }
    for(int i = 1; i <= mx; i++) sort(q[i].begin(), q[i].end()), solve(i, q[i]);    
    for(int i = 1; i <= m; i++) printf("%d\n", ans[i]);
    return 0;
}
/*
8 3
3 1 7 2 4 5 8 6 
1 6
1 3
2 4
*/