解题思路
板子题，二分那里坑了，一定要注意一下。
const int maxn = 1e5+10;
int n, m, a[maxn];
struct Node {
int s[2], v, sz, p;
void init(int _v, int _p) {
v = _v, p = _p;
sz = 1;
}
} tr[maxn*200];
int idx, rts[maxn*200];
inline void push_up(int u) {
tr[u].sz = tr[tr[u].s[0]].sz+tr[tr[u].s[1]].sz+1;
}
inline void rotate(int x) {
int y = tr[x].p, z = tr[y].p;
int k = tr[y].s[1] == x;
tr[z].s[tr[z].s[1]==y] = x, tr[x].p = z;
tr[y].s[k] = tr[x].s[k^1], tr[tr[x].s[k^1]].p = y;
tr[x].s[k^1] = y, tr[y].p = x;
push_up(y), push_up(x);
}
void splay(int &rt, int x, int k) {
while(tr[x].p!=k) {
int y = tr[x].p, z = tr[y].p;
if (z!=k)
if ((tr[y].s[1]==x)^(tr[z].s[1]==y)) rotate(x);
else rotate(y);
rotate(x);
}
if (!k) rt = x;
}
void insert(int &rt, int v) {
int u = rt, p = 0;
while(u) p = u, u = tr[u].s[v>tr[u].v];
u = ++idx;
if (p) tr[p].s[v>tr[p].v] = u;
tr[u].init(v, p);
splay(rt, u, 0);
}
void change(int &rt, int pre, int now) {
int u = rt;
//因为要保证有序性，所以不能直接修改原来的值，应该先删除再插入
while(u) {
if (pre>tr[u].v) u = tr[u].s[1];
else if (pre==tr[u].v) break;
else u = tr[u].s[0];
}
splay(rt, u, 0);
int l = tr[u].s[0], r = tr[u].s[1];
//找到原来的数的前驱和后继，后缀的左儿子就是要删的数
while(tr[l].s[1]) l = tr[l].s[1];
while(tr[r].s[0]) r = tr[r].s[0];
splay(rt, l, 0), splay(rt, r, l);
tr[r].s[0] = 0;
push_up(r), push_up(l);
insert(rt, now);
}
int get_rank(int rt, int x) {
int u = rt, sum = 0; //查询所有小于x的数的数量，+1即是排名
while(u) {
if (x>tr[u].v) {
sum += tr[tr[u].s[0]].sz+1;
u = tr[u].s[1];
}
else u = tr[u].s[0];
}
return sum;
}
int get_pre(int rt, int x) {
int u = rt, maxx = -1;
while(u) {
if (x>tr[u].v) {
maxx = max(maxx, tr[u].v);
u = tr[u].s[1];
}
else u = tr[u].s[0];
}
return maxx;
}
int get_suc(int rt, int x) {
int u = rt, minn = INF;
while(u) {
if (x<tr[u].v) {
minn = min(minn, tr[u].v);
u = tr[u].s[0];
}
else u = tr[u].s[1];
}
return minn;
}
void build(int rt, int l, int r) {
insert(rts[rt], -INF);
insert(rts[rt], INF);
for (int i = l; i<=r; ++i) insert(rts[rt], a[i]);
if (l==r) return;
int mid = (l+r)>>1;
build(rt<<1, l, mid);
build(rt<<1|1, mid+1, r);
}
void update(int rt, int l, int r, int pos, int p, int q) {
change(rts[rt], p, q);
if (l==r) return;
int mid = (l+r)>>1;
if (pos<=mid) update(rt<<1, l, mid, pos, p, q);
else update(rt<<1|1, mid+1, r, pos, p, q);
}
int query_rank(int rt, int l, int r, int L, int R, int x) {
if (l>=L && r<=R) return get_rank(rts[rt], x)-1;
int mid = (l+r)>>1, sum = 0;
if (L<=mid) sum += query_rank(rt<<1, l, mid, L, R, x);
if (R>mid) sum += query_rank(rt<<1|1, mid+1, r, L, R, x);
return sum;
}
int query_pre(int rt, int l, int r, int L, int R, int x) {
if (l>=L && r<=R) return get_pre(rts[rt], x);
int mid = (l+r)>>1, maxx = -1;
if (L<=mid) maxx = max(maxx, query_pre(rt<<1, l, mid, L, R, x));
if (R>mid) maxx = max(maxx, query_pre(rt<<1|1, mid+1, r, L, R, x));
return maxx;
}
int query_suc(int rt, int l, int r, int L, int R, int x) {
if (l>=L && r<=R) return get_suc(rts[rt], x);
int mid = (l+r)>>1, minn = INF;
if (L<=mid) minn = min(minn, query_suc(rt<<1, l, mid, L, R, x));
if (R>mid) minn = min(minn, query_suc(rt<<1|1, mid+1, r, L, R, x));
return minn;
}
int main() {
IOS;
cin >> n >> m;
for (int i = 1; i<=n; ++i) cin >> a[i];
build(1, 1, n);
int op, l, r, x;
while(m--) {
cin >> op >> l >> r;
if (op==1) {
//查询排名(从小到大数是第几个，相同看第一个)
cin >> x;
cout << query_rank(1, 1, n, l, r, x)+1 << endl;
}
else if (op==2) {
cin >> x; //查询排名的值
int L = 1, R = 10;
while(L<R) {
int mid = (L+R+1)>>1;
if (query_rank(1, 1, n, l, r, mid)+1<=x) L = mid;
else R = mid-1;
}
/*
下面的二分是错误的，比如
9 3
4 2 2 1 9 4 0 1 1
2 1 4 3
3 4 10
2 1 4 3
这组数据，第一个询问二分到3的时候，严格小于3的数有两个，算上3等于3个，但是实际上没有3这个数，
第二个询问也类似，而用上面的二分就不会有问题。
while(L<R) {
cout << L << ' ' << R << endl;
int mid = (L+R)>>1;
if (query_rank(1, 1, n, l, r, mid)+1>=x) R = mid;
else L = mid+1;
}
*/
cout << L << endl;
}
else if (op==3) {
//修改某个位置上的数
update(1, 1, n, l, a[l], r);
a[l] = r; //一定要跟新一下原来的数
}
else if (op==4) {
cin >> x; //查询前驱
cout << query_pre(1, 1, n, l, r, x) << endl;
}
else if (op==5) {
cin >> x; //查询后继
cout << query_suc(1, 1, n, l, r, x) << endl;
}
}
return 0;
}