其实思路并不算太难,就是代码量相对较大。
我们将一次离队转换为一次删除和两次加入,这样就可以保证空间是动态分配的,最大也不会暴空间。
说实话写之前感觉会很恶心,但是代码量就还好吧,有些细节需要特殊注意一下。
Code:
#include<cstdio>
using namespace std;
const int maxn = 2000000 + 5;
long long val[maxn], arr[maxn];
int f[maxn], ch[maxn][2], root[maxn], siz[maxn], nums[maxn], cnt;
struct Operator
{
inline void pushup(int x) { siz[x] = siz[ch[x][0]] + nums[x] + siz[ch[x][1]]; }
void build(int l,int r,int fa,int &cur)
{
if(l > r) return;
cur = ++ cnt;
int mid = (l + r) >> 1;
f[cur] = fa, val[cur] = arr[mid], nums[cur] = 1;
build(l,mid - 1,cur,ch[cur][0]);
build(mid + 1, r, cur, ch[cur][1]);
pushup(cur);
}
inline int get(int x){return ch[f[x]][1] == x;}
inline void rotate(int x)
{
int old = f[x], oldf = f[old], which = get(x);
ch[old][which] = ch[x][which ^ 1], f[ch[old][which]] = old;
ch[x][which ^ 1] = old, f[old] = x, f[x] = oldf;
if(oldf)ch[oldf][ch[oldf][1] == old] = x;
pushup(old);pushup(x);
}
inline void splay(int x,int & tar)
{
int a = f[tar];
for(int fa; (fa = f[x]) != a; rotate(x))
if(f[fa] != a) rotate(get(x) == get(fa) ? fa : x);
tar = x;
}
inline void insert_x(long long x,int ty,int pos,int num)
{
if(!root[ty])
{
root[ty] = ++cnt, val[root[ty]] = x, nums[root[ty]] = num;
pushup(root[ty]);
return;
}
int p = root[ty], fa, cur;
while(p)
{
fa = p;
if(pos >= siz[ch[p][0]] + nums[p]) pos -= siz[ch[p][0]] + nums[p], p = ch[p][1], cur = 1;
else p = ch[p][0], cur = 0;
}
++cnt;
val[cnt] = x, nums[cnt] = num;
ch[fa][cur] = cnt, f[cnt] = fa;
pushup(cnt);
splay(cnt,root[ty]);
}
inline int get_node(int ty,int pos)
{
int p = root[ty];
while(pos)
{
if(siz[ch[p][0]] + nums[p] < pos) { pos -= siz[ch[p][0]] + nums[p], p = ch[p][1]; continue;}
else if(pos - siz[ch[p][0]] > 0) break;
else p = ch[p][0];
}
return p;
}
inline void delete_x(int ty, int pos)
{
int p = get_node(ty,pos);
splay(p,root[ty]);
if(!ch[p][0] && !ch[p][1]) {root[ty] = 0; return; }
if(!ch[p][0]){ root[ty] = ch[p][1], f[ch[p][1]] = 0; }
else if(!ch[p][1]) { root[ty] = ch[p][0], f[ch[p][0]] = 0;}
else
{
p = ch[p][0];
while(ch[p][1]) p = ch[p][1];
splay(p,ch[root[ty]][0]);
ch[p][1] = ch[root[ty]][1], f[ch[p][1]] = p, f[p] = 0;
pushup(p);
root[ty] = p;
}
}
inline long long query(int ty,int x)
{
int p = get_node(ty,x);
splay(p,root[ty]);
return val[root[ty]] + x - siz[ch[root[ty]][0]] - 1;
}
}T;
int main()
{
//freopen("sample.txt","r",stdin);
int n,m,q;
long long k = 0;
scanf("%d%d%d",&n,&m,&q);
for(int i = 1;i <= n;++i)
{
T.insert_x(k + 1,i,0,m - 1);
k += m;
arr[i] = k;
}
T.build(1,n,0,root[n + 1]);
for(int i = 1;i <= q; ++i)
{
int x,y;
scanf("%d%d",&x,&y);
long long last = T.query(n + 1,x);
T.delete_x(n + 1,x);
if(y == m)
{
printf("%lld\n",last);
T.insert_x(last,n + 1,siz[root[n + 1]],1);
}
else
{
long long ans = T.query(x,y);
printf("%lld\n",ans);
int l = ans - val[root[x]];
int r = val[root[x]] + nums[root[x]] - 1 - ans;
int st = siz[ch[root[x]][0]] ;
T.delete_x(x,y);
if(l > 0)
{
T.insert_x(ans - l, x, st , l);
}
if(r > 0)
{
T.insert_x(ans + 1, x, st + l,r);
}
T.insert_x(ans,n + 1,siz[root[n + 1]], 1);
T.insert_x(last,x,siz[root[x]],1);
}
}
return 0;
}