bzoj5312 冒险（吉司机线段树）题解

题意：

已知\(n\)个数字，进行以下操作：

• \(1.\)区间\([L,R]\) 按位与\(x\)
• \(2.\)区间\([L,R]\) 按位或\(x\)
• \(3.\)区间\([L,R]\) 询问最大值

思路：

吉司机线段树。

我们按位考虑，维护区间或\(\_or\)和区间与\(\_and\)，那么得到区间非公有的\(1\)为\((\_or \oplus \_and)\)，那么如果对所有的非公有的\(1\)影响都一样就不会对最大值有影响，那么就直接打标机，否则继续往下更新。即

\[[(\_or[rt] \oplus \_and[rt]) \& x] == 0 || [(\_or[rt] \oplus \_and[rt]) \& x] == (\_or[rt] \oplus \_and[rt])
\]

时就直接打标记。

代码：

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 2e5 + 5;
const int MAXM = 3e6;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;

#define lson (rt << 1)
#define rson (rt << 1 | 1)
inline bool read(int &num){
    char in;
    bool IsN=false;
    in = getchar();
    if(in == EOF) return false;
    while(in != '-' && (in < '0' || in > '9')) in = getchar();
    if(in == '-'){ IsN = true; num = 0;}
    else num = in - '0';
    while(in = getchar(),in >= '0' && in <= '9'){
            num *= 10, num += in-'0';
    }
    if(IsN) num = -num;
    return true;
}

int a[maxn], all = (1 << 21) - 1;
int _or[maxn << 2], _and[maxn << 2], Max[maxn << 2];
int lazya[maxn << 2], lazyo[maxn << 2];
inline void pushup(int rt){
    _or[rt] = _or[lson] | _or[rson];
    _and[rt] = _and[lson] & _and[rson];
    Max[rt] = max(Max[lson], Max[rson]);
}
inline void pushdown(int rt, int l, int r){
    int m = (l + r) >> 1;
    if(lazya[rt] != all){
        Max[lson] &= lazya[rt];
        Max[rson] &= lazya[rt];
        _or[lson] &= lazya[rt];
        _or[rson] &= lazya[rt];
        _and[lson] &= lazya[rt];
        _and[rson] &= lazya[rt];
        lazya[lson] &= lazya[rt];
        lazya[rson] &= lazya[rt];
        lazyo[lson] &= lazya[rt];
        lazyo[rson] &= lazya[rt];
        lazya[rt] = all;
    }
    if(lazyo[rt] != 0){
        Max[lson] |= lazyo[rt];
        Max[rson] |= lazyo[rt];
        _or[lson] |= lazyo[rt];
        _or[rson] |= lazyo[rt];
        _and[lson] |= lazyo[rt];
        _and[rson] |= lazyo[rt];
        lazya[lson] |= lazyo[rt];
        lazya[rson] |= lazyo[rt];
        lazyo[lson] |= lazyo[rt];
        lazyo[rson] |= lazyo[rt];
        lazyo[rt] = 0;
    }
}
void build(int l, int r, int rt){
    lazya[rt] = all, lazyo[rt] = 0;
    if(l == r){
        _and[rt] = _or[rt] = Max[rt] = a[l];
        return;
    }
    int m = (l + r) >> 1;
    build(l, m, lson);
    build(m + 1, r, rson);
    pushup(rt);
}
void update(int L, int R, int l, int r, int op, int x, int rt){
    if(L <= l && R >= r){
        if(op == 1){    //&
            if(((_or[rt] ^ _and[rt]) & x) == 0 || ((_or[rt] ^ _and[rt]) & x) == (_or[rt] ^ _and[rt])){
                Max[rt] &= x;
                _or[rt] &= x;
                _and[rt] &= x;
                lazya[rt] &= x;
                lazyo[rt] &= x;
                return;
            }
        }
        else{   //|
            if(((_or[rt] ^ _and[rt]) & x) == 0 || ((_or[rt] ^ _and[rt]) & x) == (_or[rt] ^ _and[rt])){
                Max[rt] |= x;
                _or[rt] |= x;
                _and[rt] |= x;
                lazya[rt] |= x;
                lazyo[rt] |= x;
                return;
            }
        }
    }
    int m = (l + r) >> 1;
    pushdown(rt, l, r);
    if(L <= m)
        update(L, R, l, m, op, x, lson);
    if(R > m)
        update(L, R, m + 1, r, op, x, rson);
    pushup(rt);
}
int query(int L, int R, int l, int r, int rt){
    if(L <= l && R >= r){
        return Max[rt];
    }
    pushdown(rt, l, r);
    int m = (l + r) >> 1, MAX = -INF;
    if(L <= m)
        MAX = max(MAX, query(L, R, l, m, lson));
    if(R > m)
        MAX = max(MAX, query(L, R, m + 1, r, rson));
    return MAX;

}
int main(){
    int n, m;
    read(n), read(m);
    for(int i = 1; i <= n; i++) read(a[i]);
    build(1, n, 1);
    while(m--){
        int op, l, r, x;
        read(op), read(l), read(r);
        if(op < 3) read(x);
        if(op < 3){
            update(l, r, 1, n, op, x, 1);
        }
        else{
            printf("%d\n", query(l, r, 1, n, 1));
        }
    }
    return 0;
}