题目分析：
写的无旋treap应该跑不过，但bzoj判断的总时限。把相关实现改成线段树合并就可以了。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 1000020; 5 6 int n; 7 int ch[maxn][2],num,val[maxn]; 8 9 int son[maxn>>2][2],sz[maxn],rot[maxn],data[maxn],key[maxn]; 10 11 long long ans = 0; 12 13 int merge(int r1,int r2){ 14 if(r1 == 0) return r2; if(r2 == 0) return r1; 15 if(key[r1] < key[r2]){ 16 son[r1][1] = merge(son[r1][1],r2); 17 sz[r1] = sz[son[r1][0]]+sz[son[r1][1]]+1; 18 return r1; 19 }else{ 20 son[r2][0] = merge(r1,son[r2][0]); 21 sz[r2] = sz[son[r2][0]]+sz[son[r2][1]]+1; 22 return r2; 23 } 24 } 25 pair<int,int> split(int rt,int k){ 26 if(k == 0) return make_pair(0,rt); 27 if(k >= sz[rt]) return make_pair(rt,0); 28 if(sz[son[rt][0]] >= k){ 29 pair<int,int> res = split(son[rt][0],k); 30 son[rt][0] = res.second; res.second = rt; 31 sz[rt] = sz[son[rt][0]] + sz[son[rt][1]] + 1; 32 return res; 33 }else{ 34 pair<int,int> res = split(son[rt][1],k-sz[son[rt][0]]-1); 35 son[rt][1] = res.first; res.first = rt; 36 sz[rt] = sz[son[rt][0]] + sz[son[rt][1]] + 1; 37 return res; 38 } 39 } 40 int found(int now,int x){ 41 if(now == 0) return 0; 42 if(data[now] == x) return sz[son[now][0]]; 43 if(data[now] < x) return sz[son[now][0]]+1+found(son[now][1],x); 44 else return found(son[now][0],x); 45 } 46 47 void dfs(int now){ 48 int d; scanf("%d",&d); 49 if(d){val[now] = d; return;} 50 ch[now][0] = ++num; dfs(num); ch[now][1] = ++num; dfs(num); 51 } 52 53 long long alpha,beta; 54 void walk(int A,int B){ 55 int z = found(B,data[A]); alpha += z; beta += (sz[B]-z); 56 if(son[A][0]) walk(son[A][0],B); 57 if(son[A][1]) walk(son[A][1],B); 58 } 59 60 void dfs2(int A,int &B){ 61 if(son[A][0]) dfs2(son[A][0],B),son[A][0] = 0; 62 if(son[A][1]) dfs2(son[A][1],B),son[A][1] = 0; 63 sz[A] = 1; int z = found(B,data[A]); 64 pair<int,int> pr = split(B,z); 65 B = merge(merge(pr.first,A),pr.second); 66 } 67 68 void dfs1(int now){ 69 if(val[now]){ 70 num++; 71 key[num]=rand(); 72 sz[num]=1; 73 data[num]=val[now]; 74 rot[now]=num; 75 return; 76 } 77 dfs1(ch[now][0]); dfs1(ch[now][1]); 78 alpha = 0,beta = 0; 79 if(sz[rot[ch[now][0]]] < sz[rot[ch[now][1]]]){ 80 walk(rot[ch[now][0]],rot[ch[now][1]]); 81 dfs2(rot[ch[now][0]],rot[ch[now][1]]); 82 rot[now] = rot[ch[now][1]]; 83 }else { 84 walk(rot[ch[now][1]],rot[ch[now][0]]); 85 dfs2(rot[ch[now][1]],rot[ch[now][0]]); 86 rot[now] = rot[ch[now][0]]; 87 } 88 ans += min(alpha,beta); 89 } 90 91 void read(){ 92 scanf("%d",&n); 93 num = 1; 94 dfs(1); 95 } 96 97 void work(){ 98 num = 0; 99 dfs1(1); 100 printf("%lld",ans); 101 } 102 103 int main(){ 104 read(); 105 work(); 106 return 0; 107 }