HDU 3397 Sequence operation

题目链接

题意：给定一个01序列，有5种操作
0 a b [a.b]区间置为0
1 a b [a,b]区间置为1
2 a b [a,b]区间0变成1，1变成0
3 a b 查询[a,b]区间1的个数
4 a b 查询[a,b]区间连续1最长的长度

思路：线段树线段合并，需要两个延迟标记一个置为01，一个翻转，然后由于4操作，需要记录左边最长0、1，右边最长0、1，区间最长0、1，然后区间合并去搞即可

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define lson(x) ((x<<1)+1)
#define rson(x) ((x<<1)+2)
const int N = 100005;

int t, n, m;

struct Node {
	int l, r, sum[3][2], num;
	int setv, flip;
	Node() {
		setv = -1;
		flip = 0;
	}
	int size() {return r - l + 1;}
	void gao1(int v) {
		setv = v;
		num = v * (r - l + 1);
		for (int i = 0; i < 3; i++) {
			sum[i][v] = r - l + 1;
			sum[i][!v] = 0;
		}
		flip = 0;
	}
	void gao2() {
		flip ^= 1;
		for (int i = 0; i < 3; i++)
			swap(sum[i][0], sum[i][1]);
		num = r - l + 1 - num;
	}
} node[N * 4];

Node merge(Node lson, Node rson) {
	Node x;
	x.l = lson.l; x.r = rson.r;
	for (int i = 0; i < 2; i++) {
		x.sum[0][i] = lson.sum[0][i];
		x.sum[1][i] = rson.sum[1][i];
		x.sum[2][i] = max(lson.sum[2][i], rson.sum[2][i]);
		if (lson.sum[0][i] == lson.size())
			x.sum[0][i] += rson.sum[0][i];
		if (rson.sum[1][i] == rson.size())
			x.sum[1][i] += lson.sum[1][i];
		x.sum[2][i] = max(x.sum[2][i], lson.sum[1][i] + rson.sum[0][i]);
	}
	x.num = lson.num + rson.num;
	return x;
}

void pushup(int x) {
	node[x] = merge(node[lson(x)], node[rson(x)]);
}

void pushdown(int x) {
	if (node[x].setv != -1) {
		node[lson(x)].gao1(node[x].setv);
		node[rson(x)].gao1(node[x].setv);
		node[x].setv = -1;
	}
	if (node[x].flip) {
		node[lson(x)].gao2();
		node[rson(x)].gao2();
		node[x].flip = 0;
	}
}

void build(int l, int r, int x = 0) {
	node[x].l = l; node[x].r = r;
	node[x].setv = -1; node[x].flip = 0;
	if (l == r) {
		int tmp;
		scanf("%d", &tmp);
		for (int i = 0; i < 3; i++) {
			node[x].sum[i][tmp] = 1;
			node[x].sum[i][!tmp] = 0;
		}
		node[x].num = tmp;
		return;
	}
	int mid = (l + r) / 2;
	build(l, mid, lson(x));
	build(mid + 1, r, rson(x));
	pushup(x);
}

void add(int l, int r, int v, int x = 0) {
	if (node[x].l >= l && node[x].r <= r) {
		if (v == 2) node[x].gao2();
		else node[x].gao1(v);
		return;
	}
	int mid = (node[x].l + node[x].r) / 2;
	pushdown(x);
	if (l <= mid) add(l, r, v, lson(x));
	if (r > mid) add(l, r, v, rson(x));
	pushup(x);
}

Node query(int l, int r, int x = 0) {
	if (node[x].l >= l && node[x].r <= r)
		return node[x];
	int mid = (node[x].l + node[x].r) / 2;
	pushdown(x);
	Node ans;
	if (l <= mid && r > mid) ans = merge(query(l, r, lson(x)), query(l, r, rson(x)));
	else if (l <= mid) ans = query(l, r, lson(x));
	else if (r > mid) ans = query(l, r, rson(x));
	pushup(x);
	return ans;
}

int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &m);
		build(0, n - 1);
		int op, a, b;
		while (m--) {
			scanf("%d%d%d", &op, &a, &b);
			if (op <= 2) add(a, b, op);
			else if (op == 3) printf("%d\n", query(a, b).num);
			else if (op == 4) printf("%d\n", query(a, b).sum[2][1]);
		}
	}
	return 0;
}

HDU 3397 Sequence operation（线段树）