Lucky Queries

感覺是很簡單的區間合並， 但是好像我寫的比較麻煩。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using
 
 namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1

struct info {
    int up[2][2];
    int dn[2][2];
} a[N 
 
<< 2];

info operator + (const info& a, const info& b) {
    info ans;
    ans.up[0][0] = a.up[0][0] + b.up[0][0];
    ans.up[0][1] = max(a.up[0][0] + max(b.up[1][1], b.up[0][1]), a.up[0][1] + b.up[1][1]);
    ans.up[1][1] = a.up[1][1] + b.up[1][1];
    ans.dn[1][1] = a.dn[1][1] + b.dn[1][1
 
];
    ans.dn[1][0] = max(a.dn[1][1] + max(b.dn[0][0], b.dn[1][0]), a.dn[1][0] + b.dn[0][0]);
    ans.dn[0][0] = a.dn[0][0] + b.dn[0][0];
    return ans;
}

int lazy[N << 2];

void change(int rt) {
    swap(a[rt].up[0][0], a[rt].dn[1][1]);
    swap(a[rt].up[1][1], a[rt].dn[0][0]);
    swap(a[rt].up[0][1], a[rt].dn[1][0]);
}

void push(int rt) {
    if(lazy[rt]) {
        change(rt << 1); change(rt << 1 | 1);
        lazy[rt << 1] ^= 1;
        lazy[rt << 1 | 1] ^= 1;
        lazy[rt] = 0;
    }
}

void build(int l, int r, int rt) {
    if(l == r) {
        int x; scanf("%1d", &x);
        if(x == 4) {
            a[rt].up[0][0] = 1;
            a[rt].up[0][1] = 0;
            a[rt].up[1][1] = 0;
            a[rt].dn[0][0] = 1;
            a[rt].dn[1][0] = 0;
            a[rt].dn[1][1] = 0;
        }
        else {
            a[rt].up[0][0] = 0;
            a[rt].up[0][1] = 0;
            a[rt].up[1][1] = 1;
            a[rt].dn[0][0] = 0;
            a[rt].dn[1][0] = 0;
            a[rt].dn[1][1] = 1;
        }
        return;
    }
    int mid = l + r >> 1;
    build(lson); build(rson);
    a[rt] = a[rt << 1] + a[rt << 1 | 1];
}

void update(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) {
        lazy[rt] ^= 1;
        change(rt);
        return;
    }
    int mid = l + r >> 1;
    push(rt);
    if(L <= mid) update(L, R, lson);
    if(R > mid)  update(L, R, rson);
    a[rt] = a[rt << 1] + a[rt << 1 | 1];
}

int n, m;
char s[10];

int main() {
    scanf("%d%d", &n, &m);
    build(1, n, 1);
    while(m--) {
        scanf("%s", s);
        if(s[0] == ‘c‘) {
            printf("%d\n", max(a[1].up[0][0], max(a[1].up[0][1], a[1].up[1][1])));
        } else {
            int L, R;
            scanf("%d%d", &L, &R);
            update(L, R, 1, n, 1);
        }
    }
    return 0;
}

/*
*/

Codeforces 145E Lucky Queries 線段樹