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;}