bzoj 1858 序列操作(线段树)

题外话

本来想练练线段树的，然后发现这题及其蛋疼，要打一坨标记，这是我写过的最长的线段树了= =
然后我很SB的把R打成了r调了一个下午真是蛋疼QvQ

Description:

$给定一个0/1序列，有如下5个操作：$
$0:区间赋值为0$
$1:区间赋值为1$
$2:区间取反$
$3:询问区间内1的个数$
$4:询问区间内最大连续1的个数$

Solution

没有操作$4$这显然就是个SB题，有了操作4我们需要打几个标记
我们需要从左端点开始连续$0/1$的个数，从右端点开始的连续$0/1$个数
区间内$0/1$个数，区间内最大连续1的个数
然后我们就可以合并辣！
具体合并情况想想清楚就好了

Code

#include <bits/stdc++.h>//好题
using namespace std;
#define ls (rt << 1)
#define rs (rt << 1 | 1)
const int N = 100005;
struct Node {
    int lc[2], rc[2], mx[2], cnt[2], len; 
}p[N << 2];
int a[N], flag[N << 2];
void pushup(int rt) {
    p[rt].cnt[0] = p[ls].cnt[0] + p[rs].cnt[0];
    p[rt].cnt[1] = p[ls].cnt[1] + p[rs].cnt[1];
    p[rt].lc[0] = p[ls].lc[0] + (p[ls].lc[0] == p[ls].len ? p[rs].lc[0] : 0);
    p[rt].lc[1] = p[ls].lc[1] + (p[ls].lc[1] == p[ls].len ? p[rs].lc[1] : 0);
    p[rt].rc[0] = p[rs].rc[0] + (p[rs].rc[0] == p[rs].len ? p[ls].rc[0] : 0);
    p[rt].rc[1] = p[rs].rc[1] + (p[rs].rc[1] == p[rs].len ? p[ls].rc[1] : 0);
    p[rt].mx[0] = max(p[ls].mx[0], max(p[rs].mx[0], p[ls].rc[0] + p[rs].lc[0]));
    p[rt].mx[1] = max(p[ls].mx[1], max(p[rs].mx[1], p[ls].rc[1] + p[rs].lc[1]));
}
void build(int rt, int l, int r) {
    flag[rt] = -1;
    p[rt].len = r - l + 1;
    if (l == r) {
        p[rt].lc[a[l]] = p[rt].rc[a[l]] = p[rt].mx[a[l]] = p[rt].cnt[a[l]] = 1;
        return;
    }
    int mid = l + r >> 1;
    build(ls, l, mid);
    build(rs, mid + 1, r);
    pushup(rt);
}
void pushdown(int rt, int l, int r) {
    if (~flag[rt]) {
        int x = flag[rt];
        if (x == 2) {
            swap(p[ls].lc[0], p[ls].lc[1]), swap(p[rs].lc[0], p[rs].lc[1]);
            swap(p[ls].rc[0], p[ls].rc[1]), swap(p[rs].rc[0], p[rs].rc[1]);
            swap(p[ls].cnt[0], p[ls].cnt[1]), swap(p[rs].cnt[0], p[rs].cnt[1]);
            swap(p[ls].mx[0], p[ls].mx[1]), swap(p[rs].mx[0], p[rs].mx[1]);
            flag[ls] = 1 - flag[ls], flag[rs] = 1 - flag[rs];
        }
        else {
            p[ls].lc[x] = p[ls].rc[x] = p[ls].cnt[x] = p[ls].mx[x] = p[ls].len;
            p[ls].lc[1 - x] = p[ls].rc[1 - x] = p[ls].cnt[1 - x] = p[ls].mx[1 - x] = 0;
            p[rs].lc[x] = p[rs].rc[x] = p[rs].cnt[x] = p[rs].mx[x] = p[rs].len;
            p[rs].lc[1 - x] = p[rs].rc[1 - x] = p[rs].cnt[1 - x] = p[rs].mx[1 - x] = 0;
            flag[ls] = flag[rs] = flag[rt];
        }
        flag[rt] = -1;
    }
}
void change(int rt, int l, int r, int L, int R, int x) {
    if (L <= l && R >= r) {
        p[rt].lc[x] = p[rt].rc[x] = p[rt].cnt[x] = p[rt].mx[x] = p[rt].len;
        p[rt].lc[1 - x] = p[rt].rc[1 - x] = p[rt].cnt[1 - x] = p[rt].mx[1 - x] = 0;
        flag[rt] = x;
        return;
    }
    pushdown(rt, l, r);
    int mid = l + r >> 1;
    if (mid < L)    change(rs, mid + 1, r, L, R, x);
    else if (mid >= R)  change(ls, l, mid, L, R, x);
    else {
        change(ls, l, mid, L, mid, x);
        change(rs, mid + 1, r, mid + 1, R, x);
    }
    pushup(rt);
}
void reverse(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
        swap(p[rt].lc[0], p[rt].lc[1]), swap(p[rt].rc[0], p[rt].rc[1]);
        swap(p[rt].cnt[0], p[rt].cnt[1]), swap(p[rt].mx[0], p[rt].mx[1]);
        flag[rt] = 1 - flag[rt];
        return;
    }
    pushdown(rt, l, r);
    int mid = l + r >> 1;
    if (mid < L)    reverse(rs, mid + 1, r, L, R);
    else if (mid >= R)  reverse(ls, l, mid, L, R);
    else {
        reverse(ls, l, mid, L, mid);
        reverse(rs, mid + 1, r, mid + 1, R);
    }
    pushup(rt); 
}
int ask(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r)   return p[rt].cnt[1];
    pushdown(rt, l, r);
    int mid = l + r >> 1;
    if (mid < L)    return ask(rs, mid + 1, r, L, R);
    else if (mid >= R)  return ask(ls, l, mid, L, R);
    else return ask(ls, l, mid, L, mid) + ask(rs, mid + 1, r, mid + 1, R);
}
Node ask2(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r)   return p[rt];
    pushdown(rt, l, r);
    int mid = l + r >> 1;
    if (mid < L)    return ask2(rs, mid + 1, r, L, R);
    else if (mid >= R)  return ask2(ls, l, mid, L, R);
    else {
        Node t1 = ask2(ls, l, mid, L, mid), t2 = ask2(rs, mid + 1, r, mid + 1, R);  
        Node t3;
        t3.lc[1] = t1.lc[1] + (t1.lc[1] == t1.len ? t2.lc[1] : 0);
        t3.rc[1] = t2.rc[1] + (t2.rc[1] == t2.len ? t1.rc[1] : 0);
        t3.len = t1.len + t2.len;
        t3.mx[1] = max(t1.mx[1], max(t2.mx[1], t1.rc[1] + t2.lc[1])); 
        return t3;
    }
}
int main() {
    int n, q;
    scanf("%d%d",&n, &q);
    for (int i = 1; i <= n; ++i)    scanf("%d", &a[i]);
    build(1, 1, n);
    while (q--) {
        int x, l, r;
        scanf("%d%d%d", &x, &l, &r);
        ++l, ++r;
        switch(x) {
            case 0: change(1, 1, n, l, r, 0);break;
            case 1: change(1, 1, n, l, r, 1);break;
            case 2: reverse(1, 1, n, l, r);break;
            case 3: printf("%d\n", ask(1, 1, n, l, r));break;
            case 4: {
                Node t = ask2(1, 1, n, l, r);
                printf("%d\n", t.mx[1]);
            }
        }
    }
    return 0;
}