题目链接：T2238 礼物
考虑 $[l,r]$ 区间中哪些二元组是优秀二元组：
发现将二元组的 $x$ 按照从大到小排序，若 $y[i]$ 是这个前缀的最大值，则该二元组是优秀的.
因为数据是随机的，所以前缀最大值期望是 $log(n)$ 个的.
所以，我们可以用线段树来暴力存这些优秀的二元组.
区间合并 $pushup$ 时运用类似归并排序的方式来进行合并即可.
由于优秀二元组期望的数量是 $log$ 的，所以空间最多只会在原基础上乘一个 $log$.
我的归并写的好丑~
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#define N 200006
#define ll long long
#define mod 1000000007
#define lson t[now].ls
#define rson t[now].rs
#define siz (t[now].v.size() - 1)
#define setIO(s) freopen(s".in", "r" , stdin) , freopen(s".out", "w", stdout)
using namespace std;
int n ;
struct P
{
ll x , y;
P(ll x = 0, ll y = 0) : x(x), y(y) {}
}p[N];
struct Node
{
int ls, rs ;
vector <P> v;
}t[N << 2];
vector <P> G, V, A;
inline void pushup(int L, int R, int now)
{
if(!rson)
{
t[now].v = t[lson].v;
}
else
{
int p1 = 0, p2 = 0;
while(p1 < t[lson].v.size() && p2 < t[rson].v.size())
{
if(t[lson].v[p1].x == t[rson].v[p2].x && t[lson].v[p1].y == t[rson].v[p2].y) ++p1, ++p2;
else if(t[lson].v[p1].x >= t[rson].v[p2].x && t[lson].v[p1].y >= t[rson].v[p2].y || t[lson].v[p1].x > t[rson].v[p2].x)
{
if(t[now].v.empty() || t[lson].v[p1].y > t[now].v[t[now].v.size() - 1].y)
{
t[now].v.push_back(t[lson].v[p1]);
}
++ p1;
}
else if(t[rson].v[p2].x >= t[lson].v[p1].x && t[rson].v[p2].y >= t[lson].v[p1].y || t[rson].v[p2].x > t[lson].v[p1].x)
{
if(t[now].v.empty() || t[rson].v[p2].y > t[now].v[t[now].v.size() - 1].y)
{
t[now].v.push_back(t[rson].v[p2]);
}
++ p2;
}
}
while(p1 < t[lson].v.size())
{
if(t[now].v.empty() || (t[lson].v[p1].x <= t[now].v[siz].x && t[lson].v[p1].y > t[now].v[siz].y))
t[now].v.push_back(t[lson].v[p1]);
++p1;
}
while(p2 < t[rson].v.size())
{
if(t[now].v.empty() || (t[rson].v[p2].x <= t[now].v[siz].x && t[rson].v[p2].y > t[now].v[siz].y))
t[now].v.push_back(t[rson].v[p2]);
++p2;
}
}
}
void build(int l, int r, int now)
{
if(l == r)
{
t[now].v.push_back(P(p[l].x, p[l].y));
return ;
}
int mid = (l + r) >> 1;
build(l, mid, now << 1), t[now].ls = now << 1;
if(r > mid) build(mid + 1, r, now << 1 | 1), t[now].rs = now << 1 | 1;
pushup(l, r, now);
}
inline void query(int l, int r, int now, int L, int R)
{
if(l >= L && r <= R)
{
if(G.empty())
{
G = t[now].v;
}
else
{
V = G, A = t[now].v;
G.clear();
// merge(V, A)
int p1 = 0, p2 = 0;
while(p1 < V.size() && p2 < A.size())
{
if(V[p1].x == A[p2].x && V[p1].y == A[p2].y) ++p1, ++p2;
else if(V[p1].x >= A[p2].x && V[p1].y >= A[p2].y || V[p1].x > A[p2].x)
{
if(G.empty() || V[p1].y > G[G.size() - 1].y)
{
G.push_back(V[p1]);
}
++ p1;
}
else if(A[p2].x >= V[p1].x && A[p2].y >= V[p1].y || A[p2].x > V[p1].x)
{
if(G.empty() || A[p2].y > G[G.size() - 1].y)
{
G.push_back(A[p2]);
}
++ p2;
}
}
while(p1 < V.size())
{
if(G.empty() || (V[p1].x <= G[G.size() - 1].x && V[p1].y > G[G.size() - 1].y))
G.push_back(V[p1]);
++p1;
}
while(p2 < A.size())
{
if(G.empty() || (A[p2].x <= G[G.size() - 1].x && A[p2].y > G[G.size() - 1].y))
G.push_back(A[p2]);
++p2;
}
}
return;
}
int mid = (l + r) >> 1;
if(L <= mid) query(l, mid, now << 1, L, R);
if(R > mid) query(mid + 1, r, now << 1 | 1, L, R);
}
int main()
{
int i , j, q;
scanf("%d" , &n);
for(i = 1; i <= n ; ++i)
{
scanf("%lld%lld", &p[i].x, &p[i].y);
}
build(1, n, 1), scanf("%d", &q);
for(int cas = 1; cas <= q; ++cas)
{
int l, r;
scanf("%d%d", &l, &r);
G.clear();
query(1, n, 1, l, r);
ll w = 1;
for(j = 0; j < G.size(); ++j)
{
w *= (1ll * (ll) (G[j].x ^ G[j].y)) % mod, w %= mod;
}
printf("%lld
", w);
}
return 0;
}
卡常版：
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#define N 200006
#define ll long long
#define mod 1000000007
#define lson t[now].ls
#define rson t[now].rs
#define siz (t[now].v.size() - 1)
#define setIO(s) freopen(s".in", "r" , stdin) , freopen(s".out", "w", stdout)
using namespace std;
namespace IO
{
char *p1, *p2, buf[100000];
#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ )
int rd() {
int x = 0;
char c = nc();
while (c < 48) {
c = nc();
}
while (c > 47) {
x = (((x << 2) + x) << 1) + (c ^ 48), c = nc();
}
return x;
}
char pbuf[100000],*pp=pbuf;
void push(const char c) {
if(pp-pbuf==100000) fwrite(pbuf,1,100000,stdout),pp=pbuf;
*pp++=c;
}
void write(int x) {
static int sta[35];
int top=0;
do{sta[top++]=x%10,x/=10;}while(x);
while(top) push(sta[--top]+'0');
push('
');
}
}
int n ;
struct P
{
ll x , y;
P(ll x = 0, ll y = 0) : x(x), y(y) {}
}p[N];
struct Node
{
int ls, rs ;
vector <P> v;
}t[N << 2];
vector <P> G, V, A;
inline void pushup(int L, int R, int now)
{
if(!rson)
{
t[now].v = t[lson].v;
}
else
{
int p1 = 0, p2 = 0;
while(p1 < t[lson].v.size() && p2 < t[rson].v.size())
{
if(t[lson].v[p1].x == t[rson].v[p2].x && t[lson].v[p1].y == t[rson].v[p2].y) ++p1, ++p2;
else if(t[lson].v[p1].x >= t[rson].v[p2].x && t[lson].v[p1].y >= t[rson].v[p2].y || t[lson].v[p1].x > t[rson].v[p2].x)
{
if(t[now].v.empty() || t[lson].v[p1].y > t[now].v[t[now].v.size() - 1].y)
{
t[now].v.push_back(t[lson].v[p1]);
}
++ p1;
}
else if(t[rson].v[p2].x >= t[lson].v[p1].x && t[rson].v[p2].y >= t[lson].v[p1].y || t[rson].v[p2].x > t[lson].v[p1].x)
{
if(t[now].v.empty() || t[rson].v[p2].y > t[now].v[t[now].v.size() - 1].y)
{
t[now].v.push_back(t[rson].v[p2]);
}
++ p2;
}
}
while(p1 < t[lson].v.size())
{
if(t[now].v.empty() || (t[lson].v[p1].x <= t[now].v[siz].x && t[lson].v[p1].y > t[now].v[siz].y))
t[now].v.push_back(t[lson].v[p1]);
++p1;
}
while(p2 < t[rson].v.size())
{
if(t[now].v.empty() || (t[rson].v[p2].x <= t[now].v[siz].x && t[rson].v[p2].y > t[now].v[siz].y))
t[now].v.push_back(t[rson].v[p2]);
++p2;
}
}
}
void build(int l, int r, int now)
{
if(l == r)
{
t[now].v.push_back(P(p[l].x, p[l].y));
return ;
}
int mid = (l + r) >> 1;
build(l, mid, now << 1), t[now].ls = now << 1;
if(r > mid) build(mid + 1, r, now << 1 | 1), t[now].rs = now << 1 | 1;
pushup(l, r, now);
}
inline void query(int l, int r, int now, int L, int R)
{
if(l >= L && r <= R)
{
if(G.empty())
{
G = t[now].v;
}
else
{
V = G, A = t[now].v;
G.clear();
// merge(V, A)
int p1 = 0, p2 = 0;
while(p1 < V.size() && p2 < A.size())
{
if(V[p1].x == A[p2].x && V[p1].y == A[p2].y) ++p1, ++p2;
else if(V[p1].x >= A[p2].x && V[p1].y >= A[p2].y || V[p1].x > A[p2].x)
{
if(G.empty() || V[p1].y > G[G.size() - 1].y)
{
G.push_back(V[p1]);
}
++ p1;
}
else if(A[p2].x >= V[p1].x && A[p2].y >= V[p1].y || A[p2].x > V[p1].x)
{
if(G.empty() || A[p2].y > G[G.size() - 1].y)
{
G.push_back(A[p2]);
}
++ p2;
}
}
while(p1 < V.size())
{
if(G.empty() || (V[p1].x <= G[G.size() - 1].x && V[p1].y > G[G.size() - 1].y))
G.push_back(V[p1]);
++p1;
}
while(p2 < A.size())
{
if(G.empty() || (A[p2].x <= G[G.size() - 1].x && A[p2].y > G[G.size() - 1].y))
G.push_back(A[p2]);
++p2;
}
}
return;
}
int mid = (l + r) >> 1;
if(L <= mid) query(l, mid, now << 1, L, R);
if(R > mid) query(mid + 1, r, now << 1 | 1, L, R);
}
int main()
{
// setIO("input");
using namespace IO;
int i , j, q;
n = rd();
for(i = 1; i <= n ; ++i)
{
p[i].x = (ll) rd();
p[i].y = (ll) rd();
}
build(1, n, 1);
q = (ll) rd();
for(int cas = 1; cas <= q; ++cas)
{
int l, r;
l = rd(), r = rd();
G.clear();
query(1, n, 1, l, r);
ll w = 1;
for(j = 0; j < G.size(); ++j)
{
w *= (1ll * (ll) (G[j].x ^ G[j].y)) % mod, w %= mod;
}
write(w);
}
fwrite(pbuf,1,pp-pbuf,stdout);
return 0;
}