NOIP2017：列队

Sol

考场上：

这不是送\(50\)吗，\(Q^2\)递推就好了

然后，怎么又送\(20\)分???

\(woc\)，只有半个小时了，顺利没调出来只有\(50\)分

考后：

神\(TM\)一个大于号写成小于号。。。

\(20\)分没了

\(TAT\)

正解的一种

\(n\)棵线段树维护每一行的前\(m-1\)列

再开一棵维护最后一列的情况

长度为\(max(n, m)+q\)

动态开点

每次就变成删除节点，插入节点了

维护区间元素个数

查找就是全局第\(k\)小

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(3e5 + 5);
typedef long long ll;

IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}

int n, m, q, rt[_], num, len, tp, tail[_], now;
struct Segment{
int ls, rs, sz;
ll val;
} T[_ * 40];

IL int Size(RG int l, RG int r){
if(now == n + 1){
if(r <= n) return r - l + 1;
if(l <= n) return n - l + 1;
return 0;
}
if(r < m) return r - l + 1;
if(l < m) return m - l;
return 0;
}

IL ll Query(RG int &x, RG int l, RG int r, RG int p){
if(!x) x = ++num, T[x].sz = Size(l, r);
--T[x].sz;
if(l == r){
if(!T[x].val) T[x].val = (now == n + 1) ? 1LL * l * m : 1LL * (now - 1) * m + l;
return T[x].val;
}
RG int mid = (l + r) >> 1, sz = T[x].ls ? T[T[x].ls].sz : Size(l, mid);
if(p <= sz) return Query(T[x].ls, l, mid, p);
return Query(T[x].rs, mid + 1, r, p - sz);
}

IL void Modify(RG int &x, RG int l, RG int r, RG int p, RG ll v){
if(!x) x = ++num, T[x].sz = Size(l, r);
++T[x].sz;
if(l == r){
T[x].sz = 1, T[x].val = v;
return;
}
RG int mid = (l + r) >> 1;
if(p <= mid) Modify(T[x].ls, l, mid, p, v);
else Modify(T[x].rs, mid + 1, r, p, v);
}

int main(RG int argc, RG char* argv[]){
n = Input(), m = Input(), q = Input(), len = max(n, m) + q;
for(RG int i = 1; i <= n; ++i) T[rt[i] = ++num].sz = tail[i] = m - 1;
T[rt[n + 1] = ++num].sz = tail[n + 1] = n;
for(RG int i = 1; i <= q; ++i){
RG int x = Input(), y = Input(); RG ll id1, id2;
if(y == m){
now = n + 1, id1 = Query(rt[n + 1], 1, len, x);
++tail[n + 1], Modify(rt[n + 1], 1, len, tail[n + 1], id1);
}
else{
now = x, id1 = Query(rt[x], 1, len, y);
++tail[now = n + 1], Modify(rt[n + 1], 1, len, tail[n + 1], id1);
id2 = Query(rt[n + 1], 1, len, x);
++tail[now = x], Modify(rt[x], 1, len, tail[x], id2);
}
printf("%lld\n", id1);
}
return 0;
}