计蒜客 硬币翻转 线段树

    原题链接 https://nanti.jisuanke.com/t/11038

线段树， 每个节点保存的信息：左端点正面向上的最大长度和反面向上的最大长度，右端点正面向上的最大长度和反面向上的最大长度，该区间正面向上的最大长度和反面向上的最大长度，当然这里的最大长度是连续的。

lazy标记表示其子节点是否需要更新。

struct node{
	int l, r;   //区间[l, r] 
	int a[2];   //a[1]保存着答案 
	int L[2], R[2];
	int lazy;
}t[maxn];

AC代码

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

const int maxn = 4e5 + 5;
struct node{
	int l, r;   //区间[l, r] 
	int a[2];   //a[1]保存着答案 
	int L[2], R[2];
	int lazy;
}t[maxn];

void update(int cur){ // 更新
	node &a = t[cur];
	if(a.l == a.r) return;
	node &b = t[cur << 1], c = t[(cur << 1) + 1];
	
	//更新正面
	a.L[1] = b.L[1], a.R[1] = c.R[1]; 
	int w = b.R[1] + c.L[1];
	if(b.L[1] == b.r - b.l + 1) a.L[1] = w;
	if(c.R[1] == c.r - c.l + 1) a.R[1] = w;
	a.a[1] = max(b.a[1], c.a[1]);
	a.a[1] = max(a.a[1], w);
	
	//更新反面
	a.L[0] = b.L[0], a.R[0] = c.R[0]; 
	w = b.R[0] + c.L[0];
	if(b.L[0] == b.r - b.l + 1) a.L[0] = w;
	if(c.R[0] == c.r - c.l + 1) a.R[0] = w;
	a.a[0] = max(b.a[0], c.a[0]);
	a.a[0] = max(a.a[0], w);
		
}

void change(int cur){
	node &a = t[cur];
	swap(a.a[0], a.a[1]);
	swap(a.L[0], a.L[1]);
	swap(a.R[0], a.R[1]);
	a.lazy = 1 - a.lazy;
}

void lazy_down(int cur){ //标记下放 
	node &a = t[cur];
	int l = a.l, r = a.r;
	if(l == r || a.lazy == 0) return;
	a.lazy = 0;
	change(cur << 1);
	change((cur << 1) + 1);
}

void build(int l, int r, int cur){
	node &A = t[cur];
	A.l =l, A.r = r;
	A.lazy = 0;
	A.a[0] = A.a[1] = A.L[0] = A.L[1] = A.R[0] = A.R[1] = 0;
 	if(l == r){
		int x;
		scanf("%d", &x);
		A.L[x] = A.R[x] = A.a[x] = 1;
		return;
	}
	int mid = (l + r) / 2;
	build(l, mid, cur << 1);
	build(mid + 1, r, (cur << 1) + 1);
	update(cur);
	//printf("%d %d %d\n",l , r, t[cur].a[1]);
}

void update(int l, int r, int cur){
	int ll = t[cur].l, rr = t[cur].r;
	if(ll == l && rr == r) {
		change(cur);
		return;
	}
	lazy_down(cur);
	int mid = (ll + rr) / 2;
	if(r <= mid) update(l, r, cur << 1);
	else if(l >= mid + 1)  update(l, r, (cur << 1) + 1);
	else {
		update(l, mid, cur << 1);
		update(mid + 1, r, (cur << 1) + 1);
	}
	update(cur);
}

int GetAns(int l, int r, int cur){
	int ll = t[cur].l, rr = t[cur].r;
	if(ll == l && rr == r) return t[cur].a[1];
	lazy_down(cur);
	int mid = (ll + rr) / 2;
	if(r <= mid) return GetAns(l, r, cur << 1);
	else if(l >= mid + 1) return GetAns(l, r, (cur << 1) + 1);
	else {     //分割 
		int x = GetAns(l, mid, cur << 1);
		int y = GetAns(mid + 1, r, (cur << 1) + 1);
		x = max(x, y);
		node &b = t[cur << 1], &c = t[(cur << 1) + 1];
		int n = min(mid + 1 - l, b.R[1]);
		int m = min(r - mid, c.L[1]);
		return max(x, m + n);
	}
}


int main(){
	int n;
	while(scanf("%d", &n) == 1){
		build(1, n, 1);
		int m;
		scanf("%d", &m);
		int f, l, r;
		while(m--){
			scanf("%d%d%d", &f, &l, &r);
			if(f) update(l, r, 1);
			else printf("%d\n", GetAns(l, r, 1));
		}
	}
	return 0;
}

如有不当之处欢迎指出！