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。须要两个延迟标记一个置为01，一个翻转，然后因为4操作，须要记录左边最长0、1。右边最长0、1，区间最长0、1，然后区间合并去搞就可以

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;

}