题意：有一个长度为n的序列，由0和1组成，然后五种操作，0 a b表示把区间[a,b]内的数字全部设置为0，1 a b表示把区间[a,b]内的数字全部设置为1，2 a b表示把区间[a,b]的数字0换为1,1换为0，3 a b表示输出区间[a,b]中1的个数，4 a b表示输出区间[a,b]中连续的1的最大长度。
题解：因为要输出连续是1的最大长度，肯定合并的时候要注意。线段树维护左右端点延伸的0的数量lenl0,lenr0，还有左右端点延伸的1的数量lenl1,lenr1，和整个区间最长连续0和1的数量len0,len1，以及区间内0和1的数量num0,num1。这样操作2可以直接把所有有关0和1的维护的值全部互换。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100005;
struct Tree {
int lenl1, lenr1, len1, lenl0, lenr0, len0, num1, num0;
int flag1, flag2;
}tree[N << 2];
int n, m, A[N];
void pushup(int k, int left, int right) {
int mid = (left + right) / 2;
tree[k].lenl0 = tree[k * 2].lenl0;
tree[k].lenl1 = tree[k * 2].lenl1;
tree[k].lenr0 = tree[k * 2 + 1].lenr0;
tree[k].lenr1 = tree[k * 2 + 1].lenr1;
tree[k].num1 = tree[k * 2].num1 + tree[k * 2 + 1].num1;
tree[k].num0 = tree[k * 2].num0 + tree[k * 2 + 1].num0;
tree[k].len0 = max(tree[k * 2].lenr0 + tree[k * 2 + 1].lenl0, max(tree[k * 2].len0, tree[k * 2 + 1].len0));
tree[k].len1 = max(tree[k * 2].lenr1 + tree[k * 2 + 1].lenl1, max(tree[k * 2].len1, tree[k * 2 + 1].len1));
if (tree[k * 2].lenl0 == mid - left + 1)
tree[k].lenl0 += tree[k * 2 + 1].lenl0;
if (tree[k * 2].lenl1 == mid - left + 1)
tree[k].lenl1 += tree[k * 2 + 1].lenl1;
if (tree[k * 2 + 1]. lenr0 == right - mid)
tree[k].lenr0 += tree[k * 2].lenr0;
if (tree[k * 2 + 1]. lenr1 == right - mid)
tree[k].lenr1 += tree[k * 2].lenr1;
}
void pushdown(int k, int left, int right) {
if (tree[k].flag1 != -1) {
int mid = (left + right) / 2;
tree[k * 2].flag1 = tree[k * 2 + 1].flag1 = tree[k].flag1;
if (tree[k].flag1 == 0) {
tree[k * 2].lenl1 = tree[k * 2].lenr1 = tree[k * 2].len1 = tree[k * 2].num1 = 0;
tree[k * 2 + 1].lenl1 = tree[k * 2 + 1].lenr1 = tree[k * 2 + 1].len1 = tree[k * 2 + 1].num1 = 0;
tree[k * 2].num0 = tree[k * 2].lenl0 = tree[k * 2].lenr0 = tree[k * 2].len0 = mid - left + 1;
tree[k * 2 + 1].num0 = tree[k * 2 + 1].lenl0 = tree[k * 2 + 1].lenr0 = tree[k * 2 + 1].len0 = right - mid;
}
else if (tree[k].flag1 == 1) {
tree[k * 2].lenl0 = tree[k * 2].lenr0 = tree[k * 2].len0 = tree[k * 2].num0 = 0;
tree[k * 2 + 1].lenl0 = tree[k * 2 + 1].lenr0 = tree[k * 2 + 1].len0 = tree[k * 2 + 1].num0 = 0;
tree[k * 2].num1 = tree[k * 2].lenl1 = tree[k * 2].lenr1 = tree[k * 2].len1 = mid - left + 1;
tree[k * 2 + 1].num1 = tree[k * 2 + 1].lenl1 = tree[k * 2 + 1].lenr1 = tree[k * 2 + 1].len1 = right - mid;
}
tree[k].flag1 = -1;
tree[k * 2].flag2 = tree[k * 2 + 1].flag2 = 0;
}
if (tree[k].flag2) {
tree[k * 2].flag2 ^= tree[k].flag2;
tree[k * 2 + 1].flag2 ^= tree[k].flag2;
swap(tree[k * 2].num0, tree[k * 2].num1);
swap(tree[k * 2 + 1].num0, tree[k * 2 + 1].num1);
swap(tree[k * 2].lenl0, tree[k * 2].lenl1);
swap(tree[k * 2 + 1].lenl0, tree[k * 2 + 1].lenl1);
swap(tree[k * 2].lenr0, tree[k * 2].lenr1);
swap(tree[k * 2 + 1].lenr0, tree[k * 2 + 1].lenr1);
swap(tree[k * 2].len0, tree[k * 2].len1);
swap(tree[k * 2 + 1].len0, tree[k * 2 + 1].len1);
tree[k].flag2 = 0;
}
}
void build(int k, int left, int right) {
tree[k].flag1 = -1;
tree[k].flag2 = 0;
if (left == right) {
if (A[left] == 1) {
tree[k].num1 = tree[k].len1 = tree[k].lenl1 = tree[k].lenr1 = 1;
tree[k].num0 = tree[k].len0 = tree[k].lenl0 = tree[k].lenr0 = 0;
}
else {
tree[k].num1 = tree[k].len1 = tree[k].lenl1 = tree[k].lenr1 = 0;
tree[k].num0 = tree[k].len0 = tree[k].lenl0 = tree[k].lenr0 = 1;
}
return;
}
int mid = (left + right) / 2;
build(k * 2, left, mid);
build(k * 2 + 1, mid + 1, right);
pushup(k, left, right);
}
void modify(int k, int left, int right, int l, int r, int flag_1, int flag_2) {
if (l <= left && right <= r) {
if (flag_1 == 0) {
tree[k].flag1 = tree[k].flag2 = flag_1;
tree[k].num0 = tree[k].len0 = tree[k].lenl0 = tree[k].lenr0 = right - left + 1;
tree[k].num1 = tree[k].len1 = tree[k].lenl1 = tree[k].lenr1 = 0;
}
else if (flag_1 == 1) {
tree[k].flag1 = flag_1;
tree[k].flag2 = 0;
tree[k].num1 = tree[k].len1 = tree[k].lenl1 = tree[k].lenr1 = right - left + 1;
tree[k].num0 = tree[k].len0 = tree[k].lenl0 = tree[k].lenr0 = 0;
}
else {
tree[k].flag2 ^= flag_2;
swap(tree[k].num0, tree[k].num1);
swap(tree[k].len0, tree[k].len1);
swap(tree[k].lenl0, tree[k].lenl1);
swap(tree[k].lenr0, tree[k].lenr1);
}
return;
}
pushdown(k, left, right);
int mid = (left + right) / 2;
if (l <= mid)
modify(k * 2, left, mid, l, r, flag_1, flag_2);
if (r > mid)
modify(k * 2 + 1, mid + 1, right, l, r, flag_1, flag_2);
pushup(k, left, right);
}
int query(int k, int left, int right, int l, int r, int flag) {
if (l <= left && right <= r) {
if (flag == 1)
return tree[k].num1;
return tree[k].len1;
}
pushdown(k, left, right);
int mid = (left + right) / 2;
if (r <= mid)
return query(k * 2, left, mid, l, r, flag);
if (l > mid)
return query(k * 2 + 1, mid + 1, right, l, r, flag);
int res;
int temp1 = query(k * 2, left, mid, l, mid, flag);
int temp2 = query(k * 2 + 1, mid + 1, right, mid + 1, r, flag);
if (!flag) {
int temp3 = min(mid - l + 1, tree[k * 2].lenr1);
int temp4 = min(r - mid, tree[k * 2 + 1].lenl1);
res = max(max(temp1, temp2), temp3 + temp4);
}
else res = temp1 + temp2;
return res;
}
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++)
scanf("%d", &A[i]);
build(1, 0, n - 1);
int op, a, b;
while (m--) {
scanf("%d%d%d", &op, &a, &b);
if (op == 0)
modify(1, 0, n - 1, a, b, 0, 0);
else if (op == 1)
modify(1, 0, n - 1, a, b, 1, 0);
else if (op == 2)
modify(1, 0, n - 1, a, b, -1, 1);
else if (op == 3)
printf("%d\n", query(1, 0, n - 1, a, b, 1));
else printf("%d\n", query(1, 0, n - 1, a, b, 0));
}
}
return 0;
}
