hdu5828 Rikka with Sequence

传送门：http://acm.hdu.edu.cn/showproblem.php?pid=5828

【题解】

考虑bzoj3211 花神游历各国，只是多了区间加操作。

考虑上题写法，区间全为1打标记。考虑推广到这题：如果一个区间max开根和min开根相同，区间覆盖标记。

巧的是，这样复杂度是错的！

e.g:

$n = 10^5, m = 10^5$

$a[] = {1, 2, 1, 2, ... , 1, 2}$

$operation = { "1~1~n~2", "2~1~n", "1~1~n~2", "2~1~n", ... }$

然后发现没有可以合并的，每次都要暴力做，复杂度就错了。

考虑对于区间的$max-min leq 1$的情况维护：

当$max=min$，显然直接做即可。

当$max=min+1$，如果$sqrt{max} = sqrt{min}$，那么变成区间覆盖；否则$sqrt{max} = sqrt{min} + 1$，变成区间加法。

都是线段树基本操作，所以可以做。

下面证明复杂度为什么是对的：

时间复杂度$O(nlog^2n)$。

# include <math.h>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

# ifdef WIN32
# define LLFORMAT "%I64d"
# else
# define LLFORMAT "%lld"
# endif

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int N = 1e5 + 10;
const int mod = 1e9+7;

inline int getint() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
    return x;
}

int n, a[N];

const int SN = 262144 + 5;
struct SMT {
    ll mx[SN], mi[SN], s[SN], tag[SN], cov[SN];
    # define ls (x<<1)
    # define rs (x<<1|1)
    inline void up(int x) {
        mx[x] = max(mx[ls], mx[rs]);
        mi[x] = min(mi[ls], mi[rs]);
        s[x] = s[ls] + s[rs];
    }
    inline void pushtag(int x, int l, int r, ll tg) {
        mx[x] += tg, mi[x] += tg;
        s[x] += tg * (r-l+1); tag[x] += tg;
    }
    inline void pushcov(int x, int l, int r, ll cv) {
        mx[x] = cv, mi[x] = cv; 
        s[x] = cv * (r-l+1); cov[x] = cv; tag[x] = 0;
    }
    inline void down(int x, int l, int r) {
        register int mid = l+r>>1;
        if(cov[x]) {
            pushcov(ls, l, mid, cov[x]);
            pushcov(rs, mid+1, r, cov[x]);
            cov[x] = 0;
        }
        if(tag[x]) {
            pushtag(ls, l, mid, tag[x]);
            pushtag(rs, mid+1, r, tag[x]);
            tag[x] = 0;
        }
    }
    inline void build(int x, int l, int r) {
        tag[x] = cov[x] = 0; 
        if(l == r) {
            mx[x] = mi[x] = s[x] = a[l];
            return ;
        }
        int mid = l+r>>1;
        build(ls, l, mid); build(rs, mid+1, r);
        up(x);
    }
    inline void edt(int x, int l, int r, int L, int R, int d) {
        if(L <= l && r <= R) {
            pushtag(x, l, r, d);
            return ;
        }
        down(x, l, r);
        int mid = l+r>>1;
        if(L <= mid) edt(ls, l, mid, L, R, d);
        if(R > mid) edt(rs, mid+1, r, L, R, d);
        up(x);
    }
    inline void doit(int x, int l, int r) {
        if(mx[x] == mi[x]) {
            register ll t = mx[x];
            pushtag(x, l, r, ll(sqrt(t)) - t);
            return ;
        }
        if(mx[x] == mi[x] + 1) {
            register ll pmx = ll(sqrt(mx[x])), pmi = ll(sqrt(mi[x]));
            if(pmx == pmi) pushcov(x, l, r, pmx);
            else pushtag(x, l, r, pmx - mx[x]);    // mx[x] = mi[x] + 1
            return ;
        }
        down(x, l, r);
        int mid = l+r>>1;
        doit(ls, l, mid); doit(rs, mid+1, r);
        up(x);
    }
    
    inline void edt(int x, int l, int r, int L, int R) {
        if(L <= l && r <= R) {
            doit(x, l, r);
            return ;
        }
        down(x, l, r);
        int mid = l+r>>1;
        if(L <= mid) edt(ls, l, mid, L, R);
        if(R > mid) edt(rs, mid+1, r, L, R);
        up(x);
    }
    
    inline ll sum(int x, int l, int r, int L, int R) {
        if(L <= l && r <= R) return s[x];
        down(x, l, r);
        int mid = l+r>>1; ll ret = 0;
        if(L <= mid) ret += sum(ls, l, mid, L, R);
        if(R > mid) ret += sum(rs, mid+1, r, L, R);
        return ret;
    }
}T;

inline void sol() {
    n = getint(); register int Q = getint(), op, l, r, x;
    for (int i=1; i<=n; ++i) a[i] = getint();
    T.build(1, 1, n);
    while(Q--) {
        op = getint(), l = getint(), r = getint();
        if(op == 1) {
            x = getint();
            T.edt(1, 1, n, l, r, x);
        } else if(op == 2) T.edt(1, 1, n, l, r);
        else printf(LLFORMAT "
", T.sum(1, 1, n, l, r));
    }
}

int main() {
    int T = getint();
    while(T--) sol();
    return 0;
}