题目大意：有一个$n \times m$的方阵，第$i$行第$j$列的人的编号是$(i-1) \times m + j$。
现在有$q$个出列操作，每次让一个人出列，然后让这个人所在行向左看齐，再让最后一列向前看齐，最后让这个人站到第$n$行第$m$列的位置。
你需要输出每次出列的人的编号。
题解：可以每行维护一棵平衡树，再给最后一列维护一棵平衡树（虽然正解是用树状数组）。
发现每行的人初始编号是连续的，而对于$9 \times 10 ^ {10}$的人数，$3 \times 10 ^ {5}$次询问改变的人数其实并不多。
所以，我们把同一行内编号连续的一段缩成一个点，然后需要出列时再分裂即可。
卡点：因为正解是树状数组，所以我洛谷被玄学卡$RE$（$UOJ$过的），数组大小调了好几次，然后发现开$O(2)$，数组开的比$UOJ$大一些就不$RE$了。。。
C++ Code：（洛谷）
// luogu-judger-enable-o2
#include <cstdio>
#include <cstdlib>
#define maxn 5500010
using namespace std;
struct node{
int l, r;
long long head;
void operator -= (long long a) {l -= a; r -= a;}
} val[maxn];
int n, m, q;
int lc[maxn], rc[maxn], num[maxn], sz[maxn], tg[maxn], idx;
struct treap {
int root, len;
int ta, tb, tmp, tmp6;
int nw(node x) {
val[++idx] = x;
sz[idx] = 1;
num[idx] = rand();
lc[idx] = rc[idx] = 0;
return idx;
}
void update(int p) {sz[p] = sz[lc[p]] + sz[rc[p]] + 1;}
void pushdown(int p) {
tmp6 = tg[p]; tg[p] = 0;
val[lc[p]] -= tmp6;
val[rc[p]] -= tmp6;
tg[lc[p]] += tmp6;
tg[rc[p]] += tmp6;
}
void splitl(int rt, int k, int &x, int &y) {
if (!rt) x = y = 0;
else {
if (val[rt].l <= k) splitl(rc[rt], k, rc[rt], y), x = rt;
else splitl(lc[rt], k, x, lc[rt]), y = rt;
update(rt);
}
}
void splitr(int rt, int k, int &x, int &y) {
if (!rt) x = y = 0;
else {
pushdown(rt);
if (val[rt].r <= k) splitr(rc[rt], k, rc[rt], y), x = rt;
else splitr(lc[rt], k, x, lc[rt]), y = rt;
update(rt);
}
}
int merge(int x, int y) {
if (!x || !y) return x | y;
if (num[x] > num[y]) {
pushdown(x);
rc[x] = merge(rc[x], y);
update(x);
return x;
} else {
pushdown(y);
lc[y] = merge(x, lc[y]);
update(y);
return (y);
}
}
void build(node p) {
root = nw(p);
}
void insert(long long p, int k = -1){
if (k == -1) k = len;
if (!root) root = nw((node) {k, k, p});
else root = merge(root, nw((node) {k, k, p}));
}
long long nxt(int k) {
splitr(root, k - 1, ta, tmp);
splitl(tmp, k, tmp, tb);
int a = nw((node) {val[tmp].l, k - 1, val[tmp].head}),
b = nw((node) {k + 1, val[tmp].r, val[tmp].head + k + 1 - val[tmp].l});
tb = merge(b, tb);
tg[tb]++;
val[tb] -= 1;
root = merge(merge(ta, a), tb);
return k - val[tmp].l + val[tmp].head;
}
} s[300010], back;
int main() {
srand(20040826);
scanf("%d%d%d", &n, &m, &q);
back.len = n;
for (long long i = 1; i <= n; i++) back.insert(i * m, i);
for (long long i = 1; i <= n; i++) {
s[i].len = m - 1;
s[i].build((node){1, m - 1, (i - 1) * m + 1});
}
int x, y;
long long ta, tb;
while (q--) {
scanf("%d%d", &x, &y);
if (y != m) {
ta = s[x].nxt(y);
tb = back.nxt(x);
s[x].insert(tb);
back.insert(ta);
} else {
ta = back.nxt(x);
back.insert(ta);
}
printf("%lld\n", ta);
}
return 0;
}
C++ Code：（UOJ）
#include <cstdio>
#include <cstdlib>
#define maxn 5000010
using namespace std;
struct node{
int l, r;
long long head;
void operator -= (long long a) {l -= a; r -= a;}
} val[maxn];
int n, m, q;
int lc[maxn], rc[maxn], num[maxn], sz[maxn], tg[maxn], idx;
struct treap {
int root, len;
int ta, tb, tmp;
int nw(node x) {
val[++idx] = x;
sz[idx] = 1;
num[idx] = rand();
lc[idx] = rc[idx] = 0;
return idx;
}
void update(int p) {sz[p] = sz[lc[p]] + sz[rc[p]] + 1;}
void pushdown(int p) {
int tmp6 = tg[p]; tg[p] = 0;
val[lc[p]] -= tmp6;
val[rc[p]] -= tmp6;
tg[lc[p]] += tmp6;
tg[rc[p]] += tmp6;
}
void splitl(int rt, int k, int &x, int &y) {
if (!rt) x = y = 0;
else {
if (val[rt].l <= k) splitl(rc[rt], k, rc[rt], y), x = rt;
else splitl(lc[rt], k, x, lc[rt]), y = rt;
update(rt);
}
}
void splitr(int rt, int k, int &x, int &y) {
if (!rt) x = y = 0;
else {
pushdown(rt);
if (val[rt].r <= k) splitr(rc[rt], k, rc[rt], y), x = rt;
else splitr(lc[rt], k, x, lc[rt]), y = rt;
update(rt);
}
}
int merge(int x, int y) {
if (!x || !y) return x | y;
if (num[x] > num[y]) {
pushdown(x);
rc[x] = merge(rc[x], y);
update(x);
return x;
} else {
pushdown(y);
lc[y] = merge(x, lc[y]);
update(y);
return (y);
}
}
void build(node p) {
root = nw(p);
}
void insert(long long p, int k = -1){
if (k == -1) k = len;
if (!root) root = nw((node) {k, k, p});
else root = merge(root, nw((node) {k, k, p}));
}
long long nxt(int k) {
splitr(root, k - 1, ta, tmp);
splitl(tmp, k, tmp, tb);
int a = nw((node) {val[tmp].l, k - 1, val[tmp].head}),
b = nw((node) {k + 1, val[tmp].r, val[tmp].head + k + 1 - val[tmp].l});
tb = merge(b, tb);
tg[tb]++;
val[tb] -= 1;
root = merge(merge(ta, a), tb);
return k - val[tmp].l + val[tmp].head;
}
} s[300010], back;
int main() {
srand(20040826);
scanf("%d%d%d", &n, &m, &q);
back.len = n;
for (long long i = 1; i <= n; i++) back.insert(i * m, i);
for (long long i = 1; i <= n; i++) {
s[i].len = m - 1;
s[i].build((node){1, m - 1, (i - 1) * m + 1});
}
int x, y;
long long ta, tb;
while (q--) {
scanf("%d%d", &x, &y);
if (y != m) {
ta = s[x].nxt(y);
tb = back.nxt(x);
s[x].insert(tb);
back.insert(ta);
} else {
ta = back.nxt(x);
back.insert(ta);
}
printf("%lld\n", ta);
}
return 0;
}