线段树区间更新:
//0 x y v: [x,y]内元素都加上v
//1 x y: [x,y]内元素的和
typedef long long ll;
const int N = 100010;
struct node
{
int l, r;
ll val, mark;
}tr[N*4];
int cas = 0;
void push_up(int k)
{
tr[k].val = tr[k<<1].val + tr[k<<1|1].val;
}
void push_down(int k)
{
if(tr[k].mark)
{
tr[k<<1].mark += tr[k].mark, tr[k<<1|1].mark += tr[k].mark;
tr[k<<1].val += (tr[k<<1].r - tr[k<<1].l + 1) * tr[k].mark;
tr[k<<1|1].val += (tr[k<<1|1].r - tr[k<<1|1].l + 1) * tr[k].mark;
tr[k].mark = 0;
}
}
void build(int l, int r, int k)
{
tr[k].l = l, tr[k].r = r, tr[k].mark = 0;
if(l == r)
{
tr[k].val = 0; return;
}
int mid = (l + r) >> 1;
build(l, mid, k << 1);
build(mid + 1, r, k << 1|1);
push_up(k);
}
void update(int l, int r, int val, int k)
{
if(l <= tr[k].l && tr[k].r <= r)
{
tr[k].mark += val;
tr[k].val += 1LL * (tr[k].r - tr[k].l + 1) * val;
return;
}
push_down(k);
int mid = (tr[k].l + tr[k].r) >> 1;
if(l <= mid) update(l, r, val, k << 1);
if(r > mid) update(l, r, val, k << 1|1);
push_up(k);
}
ll query(int l, int r, int k)
{
if(l <= tr[k].l && tr[k].r <= r) return tr[k].val;
push_down(k);
int mid = (tr[k].l + tr[k].r) >> 1;
ll res = 0;
if(l <= mid) res += query(l, r, k << 1);
if(r > mid) res += query(l, r, k << 1|1);
return res;
}
int main()
{
int t, n, m, a, b, c, d;
scanf("%d", &t);
while(t--)
{
scanf("%d%d", &n, &m);
build(1, n, 1);
printf("Case %d:\n", ++cas);
for(int i = 0; i < m; i++)
{
scanf("%d", &a);
if(a == 0)
{//下标从0开始,故+1
scanf("%d%d%d", &b, &c, &d);
update(b+1, c+1, d, 1);
}
else
{
scanf("%d%d", &b, &c);
printf("%lld\n", query(b+1, c+1, 1));
}
}
}
return 0;
}
线段树区间合并:
using namespace std;
//给定一个长度为n的01串,接下来一个m,有m个操作,操作有两种:0 a b查询区间内最长连续1的长度,1 a b反转区间内,0变1,1变0
const int N = 100010;
struct node
{
int l, r;
int lone, lzero, rone, rzero, max1, max0;//分别维护左起1,左起0,右起1,右起0,区间中1的最大长度,0的最大长度
int len, mark; //区间长度,lazy标记
}tr[N*4];
int val[N];
void push_up(int k)
{
tr[k].lone = tr[k<<1].lone;
if(tr[k<<1].lone == tr[k<<1].len) tr[k].lone += tr[k<<1|1].lone;//左子树左起1的长度等于区间长度,那么直接加上右子树的左起1长度
tr[k].lzero = tr[k<<1].lzero;
if(tr[k<<1].lzero == tr[k<<1].len) tr[k].lzero += tr[k<<1|1].lzero;
tr[k].rone = tr[k<<1|1].rone;
if(tr[k<<1|1].rone == tr[k<<1|1].len) tr[k].rone += tr[k<<1].rone;
tr[k].rzero = tr[k<<1|1].rzero;
if(tr[k<<1|1].rzero == tr[k<<1|1].len) tr[k].rzero += tr[k<<1].rzero;
tr[k].max1 = max(tr[k<<1].rone + tr[k<<1|1].lone, max(tr[k<<1].max1, tr[k<<1|1].max1));
tr[k].max0 = max(tr[k<<1].rzero + tr[k<<1|1].lzero, max(tr[k<<1].max0, tr[k<<1|1].max0));
}
void push_down(int k)
{
if(tr[k].mark)//0与1互变,所以所有的维护信息也要互换
{
tr[k<<1].mark ^= tr[k].mark;
tr[k<<1|1].mark ^= tr[k].mark;
swap(tr[k<<1].lone, tr[k<<1].lzero);
swap(tr[k<<1].rone, tr[k<<1].rzero);
swap(tr[k<<1].max1, tr[k<<1].max0);
swap(tr[k<<1|1].lone, tr[k<<1|1].lzero);
swap(tr[k<<1|1].rone, tr[k<<1|1].rzero);
swap(tr[k<<1|1].max1, tr[k<<1|1].max0);
tr[k].mark = 0;
}
}
void build(int l, int r, int k)
{
tr[k].l = l, tr[k].r = r, tr[k].mark = 0, tr[k].len = r - l + 1;
if(l == r)
{
if(val[l]) tr[k].lone = tr[k].rone = tr[k].max1 = 1, tr[k].lzero = tr[k].rzero = tr[k].max0 = 0;
else tr[k].lzero = tr[k].rzero = tr[k].max0 = 1, tr[k].lone = tr[k].rone = tr[k].max1 = 0;
return;
}
int mid = (tr[k].l + tr[k].r) >> 1;
build(l, mid, k << 1);
build(mid + 1, r, k << 1 | 1);
push_up(k);
}
void update(int l, int r, int k)
{
if(l <= tr[k].l && tr[k].r <= r)
{
tr[k].mark ^= 1;
swap(tr[k].rone, tr[k].rzero);
swap(tr[k].lone, tr[k].lzero);
swap(tr[k].max1, tr[k].max0);
return;
}
push_down(k);
int mid = (tr[k].l + tr[k].r) >> 1;
if(l <= mid) update(l, r, k << 1);
if(r > mid) update(l, r, k << 1 | 1);
push_up(k);
}
int query(int l, int r, int k)
{
if(l == tr[k].l && tr[k].r == r)
return tr[k].max1;
push_down(k);
int mid = (tr[k].l + tr[k].r) >> 1;
if(r <= mid) return query(l, r, k << 1); //查询区间位于左子树
else if(l > mid) return query(l, r, k << 1 | 1); //查询区间位于右子树
else //查询区间被分割
{
int left = 0, right = 0, midd = 0;
midd = min(mid - l + 1, tr[k<<1].rone) + min(r - mid, tr[k<<1|1].lone);
left = query(l, mid, k << 1);
right = query(mid + 1, r, k << 1 | 1);
return max(midd, max(left, right));
}
}
int main()
{
int n, m, a, b, c;
while(~ scanf("%d", &n))
{
for(int i = 1; i <= n; i++)
scanf("%d", val + i);
build(1, n, 1);
scanf("%d", &m);
for(int i = 0; i < m; i++)
{
scanf("%d%d%d", &a, &b, &c);
if(a) update(b, c, 1);
else printf("%d\n", query(b, c, 1));
}
}
return 0;
}