洛谷P3960 列队 Splay
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() { 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; }