pku3667 Hotel

查询最左端的连续空房间，思想是二分，不过是在线段树上做的二分，所以线段树结点上需要记录一些附加信息，为二分提供条件

又WA了几次才过的，总结经验就是：
更新区间的过程，递归进入子节点前，如果父节点被完全覆盖（在之前的操作中整个区间被修改为住人或空房），那么要相应地修改子节点；从子节点回溯出来后，根据子节点的情况更新父节点。

查询区间的过程也是，要考虑父节点被完全覆盖的情况，这时就不用再往下递归了

#include <cstdio>

const int maxN = 50000 + 5;
struct node_t {
	int ll, mm, rr;
} tree[maxN * 3];

int max3(int x, int y, int z)
{
	x = x > y ? x : y;
	x = x > z ? x : z;
	return x;
}

int left, right;
bool checkIn;
void check(int low, int high, int node)
{
	if (left <= low && high <= right) {
		tree[node].ll = tree[node].mm = tree[node].rr = (checkIn ? 0 : high - low + 1);
	} else if (left <= high && low <= right) {
		node_t &lch = tree[node * 2], &rch = tree[node * 2 + 1];
		int mid = (low + high) / 2;
		if (tree[node].mm == high - low + 1)
		{
			lch.ll = lch.mm = lch.rr = mid - low + 1;
			rch.ll = rch.mm = rch.rr = high - mid;
		}
		if ((tree[node].ll | tree[node].mm | tree[node].rr) == 0)
		{
			lch.ll = lch.mm = lch.rr = 0;
			rch.ll = rch.mm = rch.rr = 0;
		}
		check(low, mid, node * 2);
		check(mid + 1, high, node * 2 + 1);
		tree[node].ll = lch.ll + (lch.ll == mid - low + 1 ? rch.ll : 0);
		tree[node].rr = (rch.rr == high - mid ? lch.rr : 0) + rch.rr;
		tree[node].mm = max3(lch.rr + rch.ll, lch.mm, rch.mm);
	}
}

int need;
int find(int low, int high, int node)
{
	if (tree[node].mm == high - low + 1 || (tree[node].ll | tree[node].mm | tree[node].rr) == 0) {
		if (tree[node].mm >= need)
			return low;
	} else if (low < high) {
		if (tree[node].ll >= need)
			return low;
		int mid = (low + high) / 2;
		node_t &lch = tree[node * 2], &rch = tree[node * 2 + 1];
		if (tree[node].mm >= need)
		{
			if (lch.mm >= need)
				return find(low, mid, node * 2);
			if (lch.rr + rch.ll >= need)
				return mid - lch.rr + 1;
			if (rch.mm >= need)
				return find(mid + 1, high, node * 2 + 1);
		}
		if (tree[node].rr >= need)
			return high - tree[node].rr + 1;
	}
	return 0;
}

int main()
{
	int N, M, req, X, D;
	scanf("%d%d", &N, &M);
	left = 1, right = N, checkIn = false;
	check(1, N, 1);
	while (M--)
	{
		scanf("%d", &req);
		if (req == 1) {
			scanf("%d", &D);
			need = D;
			X = find(1, N, 1);
			printf("%d\n", X);
			if (X != 0)
			{
				left = X, right = X + D - 1, checkIn = true;
				check(1, N, 1);
			}
		} else {
			scanf("%d%d", &X, &D);
			left = X, right = X + D - 1, checkIn = false;
			check(1, N, 1);
		}
	}
	return 0;
}