学步园

两个题目都是用同一个模板，询问最长的连续未覆盖的区间 。

lazy代表是否有人，msum代表区间内最大的连续长度，lsum是从左结点往右的连续长度，rsum是从右结点往左的连续长度。

区间合并很恶心啊，各种左左右右左右左右........

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
# define MAX 211111
# define ll(x) x << 1
# define rr(x) x << 1 | 1
using namespace std;
struct node {
    int l,r,mid,lazy;
    int msum,lsum,rsum;

} tree[MAX*4];

void up(int x) {
    int p = tree[x].r - tree[x].l + 1;
    tree[x].msum = max(tree[rr(x)].lsum + tree[ll(x)].rsum, max(tree[ll(x)].msum, tree[rr(x)].msum));
    tree[x].lsum = tree[ll(x)].lsum;
    tree[x].rsum = tree[rr(x)].rsum;
    if(tree[x].lsum == p - (p >> 1)) tree[x].lsum += tree[rr(x)].lsum;
    if(tree[x].rsum == (p >> 1)) tree[x].rsum += tree[ll(x)].rsum;
}
void down (int x) {
    int p = tree[x].r - tree[x].l + 1;
    if(tree[x].lazy != -1) {
        tree[ll(x)].lazy = tree[rr(x)].lazy = tree[x].lazy;
        tree[ll(x)].lsum = tree[ll(x)].rsum = tree[ll(x)].msum = tree[x].lazy ? 0 : p - (p>>1);
        tree[rr(x)].lsum = tree[rr(x)].rsum = tree[rr(x)].msum = tree[x].lazy ? 0 : (p>>1);
        tree[x].lazy = -1;
    }
}

void build(int l,int r,int x) {
    tree[x].l = l;
    tree[x].r = r;
    tree[x].mid = (l+r) >> 1;
    tree[x].lazy = -1;
    tree[x].lsum = tree[x].rsum = tree[x].msum = r - l + 1;
    if(l == r) {
        return ;
    }
    build(l,tree[x].mid,ll(x));
    build(tree[x].mid+1,r,rr(x));
}

void update(int l,int r,int v,int x) {
    if(l == tree[x].l && r == tree[x].r) {
        tree[x].msum = tree[x].lsum = tree[x].rsum = v ? 0: tree[x].r - tree[x].l + 1;
        tree[x].lazy = v;
        return ;
    }
    down(x);
    if(r <= tree[x].mid ) update(l,r,v,ll(x));
    else if(l > tree[x].mid) update(l,r,v,rr(x));
    else {
        update(l,tree[x].mid,v,ll(x));
        update(tree[x].mid + 1,r,v,rr(x));
    }
    up(x);
}

int query(int v,int x) {
    if(tree[x].l == tree[x].r) return tree[x].l;
    down(x);
    if(tree[ll(x)].msum >= v) return query(v,ll(x));
    else if(tree[ll(x)].rsum + tree[rr(x)].lsum >= v) return tree[x].mid - tree[ll(x)].rsum + 1;
    else {
        return query(v,rr(x));
    }
}


int main() {
   

    return 0;
}