版权声明：https://blog.csdn.net/qq_41730082 https://blog.csdn.net/qq_41730082/article/details/88204248
题目链接
好题！至少在线段树上面又有了新的想法。
一开始的做法是写了Splay树，然后答案就是找合法后继，但是时间复杂度略高（TLE在第30组）——后面有相应的代码附上（想看的可以看一下哦，当作学习Splay还是不错的），之后就是一直在不断的进行优化，然后过程之中，想到了既然要维护二维图上符合这个条件的下一个节点，不妨就去离散化，然后去寻找这样的符合条件的节点。
这么应该怎么去写，一开始的Splay倒是给了我一点思想，当时造出来的一组数据提点到了我，我们既然要找下一个符合条件的点，要满足x、y均要强制大于它，所以我们似乎可以去建一棵线段树来维护，怎么维护呢？就是要用到了set的一些性质，我们去得到每个数的其中的最大的y的坐标维护进线段树中，然后我们所要找的[l, r]的区间，就只要去维护x的关系就行了，需要满足的是r > x并且还有该段区间的val > y这两个条件即可。
首先建树，然后维护区间最大值。
测试样例：
10
a 1 4
a 2 3
a 3 5
a 3 7
f 1 4
f 1 2
r 3 5
a 4 3
f 1 4
f 3 2
Output:
3 5
2 3
3 7
4 3AC代码：线段树
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 2e5 + 7;
int N, lisn[maxN], tot, diff, tree[maxN<<2];
set<int> st[maxN];
struct Ques
{
char op[10];
int x, y;
}q[maxN];
void buildTree(int rt, int l, int r)
{
tree[rt] = -1;
if(l == r) return;
int mid = HalF;
buildTree(Lson);
buildTree(Rson);
}
void update(int rt, int l, int r, int qx)
{
if(l == r)
{
if(st[qx].size()) tree[rt] = *(-- st[qx].end());
else tree[rt] = -1;
return;
}
int mid = HalF;
if(qx <= mid) update(Lson, qx);
else update(Rson, qx);
tree[rt] = max(tree[lsn], tree[rsn]);
}
int query(int rt, int l, int r, int qx, int qy)
{
if(r <= qx || tree[rt] <= qy) return -1;
if(l == r) return l;
int mid = HalF;
int ans = query(Lson, qx, qy);
if(ans == -1) ans = query(Rson, qx, qy);
return ans;
}
inline void init()
{
tot = 0;
}
int main()
{
scanf("%d", &N);
init();
for(int i=1; i<=N; i++)
{
scanf("%s%d%d", q[i].op, &q[i].x, &q[i].y);
lisn[++tot] = q[i].x;
}
sort(lisn + 1, lisn + tot + 1);
diff = (int)( unique(lisn + 1, lisn + tot + 1) - lisn - 1 );
buildTree(1, 1, diff);
for(int i=1, x, y, ans; i<=N; i++)
{
x = (int)(lower_bound(lisn + 1, lisn + diff + 1, q[i].x) - lisn);
y = q[i].y;
if(q[i].op[0] == 'a')
{
st[x].insert(y);
update(1, 1, diff, x);
}
else if(q[i].op[0] == 'r')
{
st[x].erase(y);
update(1, 1, diff, x);
}
else
{
ans = query(1, 1, diff, x, y);
if(ans ^ -1) printf("%d %d\n", lisn[ans], *st[ans].upper_bound(y));
else printf("-1\n");
}
}
return 0;
}
TLE代码：又长又复杂但是想了我好久的Splay。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 2e5 + 7;
int N, l_x[maxN], l_y[maxN], t_x, t_y, dx, dy, root, cnt;
set<int> st[maxN<<1];
set<int>::iterator it;
struct Ques
{
char op[10];
int x, y;
}q[maxN];
struct node
{
int ff, val, cnt, size, ch[2];
node() { ff = val = cnt = size = ch[0] = ch[1] = 0; }
void init(int fa, int _val) { ff = fa; val = _val; cnt = size = 1; ch[0] = ch[1] = 0; }
}t[maxN<<3];
void pushup(int rt) { t[rt].size = t[t[rt].ch[0]].size + t[t[rt].ch[1]].size + t[rt].cnt; }
void Rotate(int x)
{
int y = t[x].ff, z = t[y].ff;
int k = t[y].ch[1] == x;
t[z].ch[t[z].ch[1] == y] = x;
t[x].ff = z;
t[y].ch[k] = t[x].ch[k^1];
t[t[x].ch[k^1]].ff = y;
t[x].ch[k^1] = y;
t[y].ff = x;
pushup(y); pushup(x);
}
void Splay(int x, int goal)
{
while(t[x].ff != goal)
{
int y = t[x].ff, z = t[y].ff;
if(z != goal) (t[z].ch[0] == y) ^ (t[y].ch[0] == x) ? Rotate(x) : Rotate(y);
Rotate(x);
}
if(!goal) root = x;
}
void insert(int x)
{
int u = root, ff = 0;
while(u && t[u].val != x)
{
ff = u;
u = t[u].ch[x > t[u].val];
}
if(u) t[u].cnt++;
else
{
u = ++cnt;
if(ff) t[ff].ch[x > t[ff].val] = u;
t[u].init(ff, x);
}
Splay(u, 0);
}
void Find(int x)
{
int u = root;
if(!u) return;
while(t[u].ch[x > t[u].val] && x != t[u].val) u = t[u].ch[x > t[u].val];
Splay(u, 0);
}
int Next(int x, int f)
{
Find(x);
int u = root;
if(t[u].val > x && f) return u;
if(t[u].val < x && !f) return u;
u = t[u].ch[f];
while(t[u].ch[f^1]) u = t[u].ch[f^1];
return u;
}
void Delet(int x)
{
int last = Next(x, 0), next = Next(x, 1);
Splay(last, 0); Splay(next, last);
int del = t[next].ch[0];
if(t[del].cnt > 1) { t[del].cnt--; Splay(del, 0); }
else t[next].ch[0] = 0;
}
inline void init()
{
t_x = t_y = root = cnt = 0;
insert(INF);
insert(-INF);
}
int main()
{
scanf("%d", &N);
init();
for(int i=1; i<=N; i++)
{
scanf("%s%d%d", q[i].op, &q[i].x, &q[i].y);
l_x[++t_x] = q[i].x; l_y[++t_y] = q[i].y;
}
sort(l_x + 1, l_x + t_x + 1);
sort(l_y + 1, l_y + t_y + 1);
dx = (int)( unique(l_x + 1, l_x + t_x + 1) - l_x - 1 );
dy = (int)( unique(l_y + 1, l_y + t_y + 1) - l_y - 1 );
for(int i=1, x, y; i<=N; i++)
{
x = (int)( lower_bound(l_x + 1, l_x + dx + 1, q[i].x) - l_x );
y = (int)( lower_bound(l_y + 1, l_y + dy + 1, q[i].y) - l_y );
if(q[i].op[0] == 'a') //add
{
insert(x);
st[x].insert(y);
}
else if(q[i].op[0] == 'r') //remove
{
st[x].erase(y);
Delet(x);
}
else //find
{
while(true)
{
int nex = Next(x, 1), tmp = t[nex].val;
if(tmp == INF) { printf("-1\n"); break; }
it = st[tmp].upper_bound(y);
if(it == st[tmp].end()) { x = tmp; continue; }
printf("%d %d\n", l_x[tmp], l_y[*it]);
break;
}
}
}
return 0;
}