分块
tot = cnt = 1;
d = sqrt(n);
for (int i = 1; i <= n; i++)
{
vis[i] = 0;
}
for (int i = 1; i <= n; i++)
{
blocks[i] = cnt;
if (tot == d)
{
tot = 1;
cnt++;
}
else
{
tot++;
}
}
for (int i = 1; i <= cnt; i++)
{
blockvis[i] = blocksum[i] = blocksize[i] = 0;
}
for (int i = 1; i <= n; i++)
{
blocksize[blocks[i]]++;
}
for (ll i = 1; i <= m; i++)
{
if (blocks[nodes[i].l] == blocks[nodes[i].r])
{
if (blockvis[blocks[nodes[i].l]] != blocksize[blocks[nodes[i].l]])
{
for (ll j = nodes[i].l; j <= nodes[i].r; j++)
{
}
}
}
else
{
if (blockvis[blocks[nodes[i].l]] != blocksize[blocks[nodes[i].l]])
{
for (ll j = nodes[i].l; blocks[j] == blocks[nodes[i].l]; j++)
{
}
}
if (blockvis[blocks[nodes[i].r]] != blocksize[blocks[nodes[i].r]])
{
for (ll j = nodes[i].r; blocks[j] == blocks[nodes[i].r]; j--)
{
}
}
for (ll j = blocks[nodes[i].l] + 1; j < blocks[nodes[i].r]; j++)
{
}
}
}
ll t, ans;
ll blo[maxn];
int main()
{
// DEBUG;
scanf("%lld", &t);
blo[0] = 5;
ll block = 1e4;
for (ll i = 1; i <= 1e5; i++)
{
blo[i] = (i + 1) * block;
}
while (t--)
{
ll L, R;
scanf("%lld%lld", &L, &R);
ll l = L / block + 1, r = R / block - 1;
ll minn = INF;
if (R - L <= 5 * block)
{
for (ll i = L; i <= R; i++)
{
}
continue;
}
for (ll i = L; i <= l * block; i++)
{
}
for (ll i = r * block + 1; i <= R; i++)
{
}
// printf("*1\n");
for (ll i = l; i <= r; i++)
{
}
}
}
莫队
T(n): O((n+m)*sqrt(n))
ll n,q,blo,curl,curr,cnt[maxn],answer,ans[maxn];
ll arr[maxn];
struct Query
{
ll l,r,pos;
bool operator<(Query other)const
{
if(l/blo!=other.l/blo) return l<other.l;
else return r<other.r;
}
}qry[maxn];
void add(ll pos)
{
cnt[arr[pos]]++;
if(cnt[arr[pos]]==1) answer++;//每个题都不一样
}
void del(ll pos)
{
cnt[arr[pos]]--;
if(cnt[arr[pos]]==0) answer--;//每个题都不一样
}
int main()
{
scanf("%lld%lld%lld", &n, &q, &k);
blo = sqrt(n);
for (ll i = 1; i <= n; i++)
{
scanf("%lld", &arr[i]);
}
for (ll i = 1; i <= q; i++)
{
ll l, r;
scanf("%lld%lld", &l, &r);
qry[i] = {
l, r, i};
}
sort(qry + 1, qry + q + 1);
curl=curr=1;
cnt[arr[1]]++;
sum+=cnt[arr[1]]*cnt[arr[1]];
for (ll i = 1; i <= q; i++)
{
ll l = qry[i].l, r = qry[i].r;
while (curl < l)
del(curl++);
while (curl > l)
add(--curl);
while (curr < r)
add(++curr);
while (curr > r)
del(curr--);
ans[qry[i].pos] = sum;
}
for (ll i = 1; i <= q; i++)
{
printf("%lld\n", ans[i]);
}
}
带修莫队
T(n): O(n^(3/5))
ll n, q, blo, curl, curr, cnt[maxn], answer, ans[maxn], sum, k, xiu,now,qnum;
ll arr[maxn];
struct Query
{
ll l, r, pos, xiu;
bool operator<(Query other) const
{
if (l / blo != other.l / blo)
return l < other.l;
else if(r / blo != other.r / blo)
return r < other.r;
else return xiu<other.xiu;
}
} qry[maxn];
struct Xiu
{
ll pos,val;
}xs[maxn];
void add(ll pos)
{
cnt[arr[pos]]++;
if(cnt[arr[pos]]==1) sum++;
}
void del(ll pos)
{
cnt[arr[pos]]--;
if(cnt[arr[pos]]==0) sum--;
}
void work(ll now,ll l,ll r,ll x)
{
if(xs[now].pos>=l&&xs[now].pos<=r)
{
cnt[arr[xs[now].pos]]--;
if(cnt[arr[xs[now].pos]]==0) sum--;
cnt[xs[now].val]++;
if(cnt[xs[now].val]==1) sum++;
}
swap(arr[xs[now].pos],xs[now].val);
}
int main()
{
// DEBUG;
scanf("%lld%lld", &n, &q);
blo = pow(n,2.0/3);
for (ll i = 1; i <= n; i++)
{
scanf("%lld", &arr[i]);
}
for (ll i = 1; i <= q; i++)
{
char op[20];
scanf("%s", op);
if (op[0] == 'Q')
{
ll l, r;
scanf("%lld%lld", &l, &r);
qry[++qnum] = {
l, r, qnum,xiu};
}
else
{
ll x,c;
scanf("%lld%lld", &x, &c);
xs[++xiu]={
x,c};
}
}
sort(qry + 1, qry + qnum + 1);
curl = curr = 1,now=0;
cnt[arr[1]]++;
sum += cnt[arr[1]] * cnt[arr[1]];
for (ll i = 1; i <= q; i++)
{
ll l = qry[i].l, r = qry[i].r,x=qry[i].xiu;
while (curl < l)
del(curl++);
while (curl > l)
add(--curl);
while (curr < r)
add(++curr);
while (curr > r)
del(curr--);
while(now<x)
work(++now,l,r,x);
while(now>x)
work(now--,l,r,x);
ans[qry[i].pos]=sum;
}
for (ll i = 1; i <= qnum; i++)
{
printf("%lld\n", ans[i]);
}
}