真是比出纳员还郁闷...
splay有点无法理解 可能是被机房dalao影响心态了吧
嘤嘤嘤还是Treap好写
这题主要难度在于批量删除并统计答案
发现其实删除子树还是很方便的
统计答案我就删一个ans++
其实可以统计插入了几个和最后剩几个这样应该能方便点
Code:
1 #include<cstdio> 2 #include<iostream> 3 #include<cstdlib> 4 #include<ctime> 5 #include<cstring> 6 #include<vector> 7 #include<cmath> 8 #include<algorithm> 9 #define rep(i,a,n) for(int i = a;i <= n;i++) 10 #define per(i,n,a) for(int i = n;i >= a;i--) 11 using namespace std; 12 typedef long long ll; 13 int read() { 14 int as = 0,fu = 1; 15 char c = getchar(); 16 while(c < '0' || c > '9') { 17 if(c == '-') fu = -1; 18 c = getchar(); 19 } 20 while(c >= '0' && c <= '9') { 21 as = as * 10 + c - '0'; 22 c = getchar(); 23 } 24 return as * fu; 25 } 26 const int N = 300005; 27 //head 28 int root,ans,tot; 29 int val[N],key[N],sze[N],sn[2][N]; 30 #define ls sn[0][x] 31 #define rs sn[1][x] 32 inline void pup(int x) {if(x) sze[x] = sze[ls] + sze[rs] + 1;} 33 34 inline void Rot(int& x,int d) { 35 int tmp = sn[d^1][x]; 36 sn[d^1][x] = sn[d][tmp]; 37 sn[d][tmp] = x; 38 pup(x),pup(tmp); 39 x = tmp; 40 } 41 42 void Ins(int &x,int v) { 43 if(!x) { 44 x = ++tot; 45 sze[x] = 1,val[x] = v,key[x] = rand(); 46 } else { 47 bool d = v > val[x]; 48 Ins(sn[d][x],v); 49 if(key[sn[d][x]] > key[x]) Rot(x,d^1); 50 } 51 pup(x); 52 } 53 54 void Del(int &x,int v) { 55 if(val[x] == v) { 56 ans++; 57 if(!ls || !rs) return void(x = ls|rs); 58 if(val[ls] > val[rs]) Rot(x,0),Del(ls,v); 59 else Rot(x,1),Del(rs,v); 60 } else Del(sn[v > val[x]][x],v); 61 pup(x); 62 } 63 64 void Delt(int &x) { 65 if(!x) return; 66 Delt(ls),Delt(rs); 67 pup(x),x = 0,ans++; 68 } 69 70 void Dlowerbound(int &x,int v) { 71 if(!x) return; 72 if(v > val[x]) { 73 Delt(ls),Dlowerbound(rs,v); 74 Del(x,val[x]); 75 } else Dlowerbound(ls,v); 76 pup(x); 77 } 78 79 int rnkq(int x,int rnk) { 80 if(rnk <= sze[ls]) return rnkq(ls,rnk); 81 if(rnk <= sze[ls] + 1) return val[x]; 82 return rnkq(rs,rnk - sze[ls] - 1); 83 } 84 int n,x,minn,delta; 85 char cmd[10]; 86 int main() { 87 n = read(),minn = read(); 88 rep(i,1,n) { 89 scanf("%s%d",cmd,&x); 90 if(cmd[0] == 'I') if(x >= minn) Ins(root,x - delta); 91 if(cmd[0] == 'A') delta += x; 92 if(cmd[0] == 'S') { 93 delta -= x; 94 int tmp = minn - delta; 95 Dlowerbound(root,tmp); 96 } 97 if(cmd[0] == 'F') { 98 if(sze[root] < x) puts("-1"); 99 else printf("%d ",rnkq(root,sze[root] - x + 1) + delta); 100 } 101 } 102 printf("%d ",ans); 103 return 0; 104 }