c/c++语言开发共享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("%dn", ans[i]);     return 0; } /* 8 3 3 1 7 2 4 5 8 6  1 6 1 3 2 4 */