【NOI2004】【splay】【SBT】郁闷的出纳员

这道题用很多数据结构都可以做，这里用splay实现。

因为增加工资和减少工资都是对所有员工进行操作，所以维护一个delta就行，因为操作只对之前的工资档案有效，所以在新加入数据时需先减去delta。

还有就是如果刚来就低于下界的人是不计入离开的总人数的。

splay维护子树的size以及每个数据的个数cnt，其他就是基本的操作。

代码：

#include<cstdio>#include<cstring>using namespace std;const int inf = 0x3f3f3f3f;const int maxn = 100000 + 10;int delta = 0,ans = 0;int size = 0,root = 0;int n,min_cost;struct SplayTree{int data[maxn];int cnt[maxn];int ch[maxn][2],t_size[maxn][2];int pre[maxn];inline int min(int a,int b){return a < b ? a : b;}inline void rotate(int x,int f){int y = pre[x],z = pre[y];ch[y][!f]= ch[x][f];t_size[y][!f] = t_size[x][f];pre[ ch[x][f] ] = y;ch[x][f] = y;t_size[x][f] = t_size[y][0] + t_size[y][1] + cnt[y];pre[y] = x;pre[x] = z;if(z){ch[z][ch[z][1] == y] = x;t_size[z][ch[z][1] == x] = t_size[x][0] + t_size[x][1] + cnt[x];}}void splay(int x){int y,z;while(pre[x] != 0){y = pre[x],z = pre[y];if(z == 0){if(x == ch[y][0])rotate(x,1);else rotate(x,0);break;}if(x == ch[y][0]){if(y == ch[z][0])rotate(y,1),rotate(x,1);else rotate(x,1),rotate(x,0);}else{if(y == ch[z][1])rotate(y,0),rotate(x,0);else rotate(x,0),rotate(x,1);}}root = x;}void insert(int x){data[++size] = x;cnt[size] = 1;if(root == 0){root = size;return;}int pos = root,tmp;while(pos != 0){tmp = pos;if(x == data[tmp]){cnt[tmp]++;splay(tmp);return;}if(x < data[tmp])pos = ch[pos][0];else pos = ch[pos][1];}pre[size] = tmp;if(x <= data[tmp])ch[tmp][0] = size;else ch[tmp][1] = size;splay(size);}void del(){int p = root,tmp = -1;while(p != 0){if(data[p] + delta < min_cost)tmp = p,p = ch[p][1];else p = ch[p][0];}if(tmp == -1)return;p = tmp;splay(p);ans += t_size[p][0]  + cnt[p];root = ch[p][1];if(root != 0)pre[root] = 0;}int query(int k,int p){if(t_size[p][1] < k && t_size[p][1] + cnt[p] >= k)return data[p];if(t_size[p][1] >= k)return query(k,ch[p][1]);else return query(k - t_size[p][1] - cnt[p],ch[p][0]);}}Spt;void init(){freopen("bzoj1503.in","r",stdin);freopen("bzoj1503.out","w",stdout);}void readdata(){scanf("%d%d",&n,&min_cost);}void solve(){for(int i = 1;i <= n;i++){char op[2];int p;scanf("%s%d",op,&p);if(op[0] == 'I'){if(p >= min_cost){p -= delta;Spt.insert(p);}}if(op[0] == 'A')delta += p;if(op[0] == 'S'){delta -= p;Spt.del();}if(op[0] == 'F'){if(p > Spt.t_size[root][0] + Spt.t_size[root][1] + Spt.cnt[root])printf("-1\n");else printf("%d\n",Spt.query(p,root) + delta);}}printf("%d",ans);}int main(){init();readdata();solve();return 0;}

然后另一个是使用SBT实现的，总的来说SBT比splay实现的平衡树效率要高很多。

代码:

#include<cstdio>#include<cstring>#define L ch[p][0]#define R ch[p][1]#define LL ch[ch[p][0]][0]#define RR ch[ch[p][1]][1]#define LR ch[ch[p][0]][1]#define RL ch[ch[p][1]][0]using namespace std;const int maxn = 200000;int ch[maxn][2];int sz[maxn],val[maxn];int n,min,root;int top = 0,delta = 0;int ans = 0;void init(){freopen("cashier.in","r",stdin);freopen("cashier.out","w",stdout);}void Rotate(int &p,int f){int k = ch[p][!f];ch[p][!f] = ch[k][f];ch[k][f] = p;sz[k] = sz[p];sz[p] = sz[L] + sz[R] + 1;p = k;}void maintain(int &p,bool flag){if(p == 0)return;if(!flag){if(sz[LL] > sz[R])Rotate(p,1);else if(sz[LR] > sz[R])Rotate(L,0),Rotate(p,1);else return;}else{if(sz[RR] > sz[L])Rotate(p,0);else if(sz[RL] > sz[L])Rotate(R,1),Rotate(p,0);else return;}maintain(L,false);maintain(R,true);maintain(p,false);maintain(p,true);}void insert(int &p,int key){if(p == 0){p = ++top;ch[p][0] = ch[p][1] = 0;sz[p] = 1;val[p] = key;return;}sz[p]++;if(key < val[p])insert(ch[p][0],key);else insert(ch[p][1],key);maintain(p,!(key < val[p]));}void del(int &p){if(!p)return;if(val[p] + delta < min)p = ch[p][1],del(p);else del(ch[p][0]),sz[p] = sz[ ch[p][0] ] + sz[ ch[p][1] ] + 1;}int find(int &p,int k){if(sz[ ch[p][1] ] >= k)return find(ch[p][1],k);else if(sz[ ch[p][1] ] + 1 == k)return val[p]; else return find(ch[p][0],k - sz[ ch[p][1] ] - 1);}void readdata(){scanf("%d%d",&n,&min);int tot = 0;for(int i = 1;i <= n;i++){char op[2];int tmp;scanf("%s%d",op,&tmp);if(op[0] == 'I'){if(tmp >= min)insert(root,tmp - delta),tot++;}if(op[0] == 'A')delta += tmp;if(op[0] == 'S'){delta -= tmp;del(root);}if(op[0] == 'F'){if(sz[root] < tmp)printf("-1\n");else printf("%d\n",find(root,tmp) + delta);}}printf("%d\n",tot - sz[root]);}int main(){init();readdata();return 0;}