BZOJ 1503 郁闷的出纳员【NOI2004】

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1503

很明显，因为修改工资是对所有人操作的，因此我们可以对于每一个新员工，先减去之前的修改，就可以认为他与前面的人经过了相同的修改，再将修改后的值插入Splay即可。有一点特别坑：题目上明明说刚来就嫌待遇太差(100块都不给我！)的算离开的员工，但答案中并没有统计这种情况。。。害得我WA了好久。。。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int inf = 0x3f3f3f3f;

struct Node {
int val,sz,num;
Node *fa,*ch[2];
}*root,*null,*L,*R,T[100000+10];

int n,minn,ans,k,cnt;

Node* Newnode(Node *f,int val) {
T[cnt].fa = f;T[cnt].val = val;T[cnt].sz = 0;T[cnt].ch[0] = T[cnt].ch[1] = null;T[cnt].num = 1;return &T[cnt++];
}

void Update(Node *u) {
u -> sz = u -> num;
if(u -> ch[0] != null)u -> sz += u -> ch[0] -> sz;
if(u -> ch[1] != null)u -> sz += u -> ch[1] -> sz;
}

void Rotate(Node *u) {
Node *x = u -> fa,*y = x -> fa;int v = u == x -> ch[1];
u -> fa = y;if(y != null)y -> ch[x == y -> ch[1]] = u;
x -> ch[v] = u -> ch[v^1];if(u -> ch[v^1] != null)u -> ch[v^1] -> fa = x;
x -> fa = u;u -> ch[v^1] = x;Update(x);
}

void Splay(Node *u,Node *f) {
Node *x;
for(;(x = u -> fa) != f;Rotate(u))if(x -> fa != f)Rotate((u == x -> ch[1]) == (x == x -> fa -> ch[1]) ? x : u);
Update(u);if(f == null)root = u;
}

void init() {
null = Newnode(null,0);L = root = Newnode(null,-inf);R = root -> ch[1] = Newnode(root,inf);Update(root -> ch[1]);Update(root);
}

void Insert(int x) {
Node *u = root;int t;
while(u != null) {
if(u -> val == x) {u -> num++;Splay(u,null);return;}
if(x < u -> val) {
if(u -> ch[0] == null) {
Splay((u -> ch[0] = Newnode(u,x)),null);return;
}
u = u -> ch[0];
}
else {
if(u -> ch[1] == null) {
Splay((u -> ch[1] = Newnode(u,x)),null);return;
}
u = u -> ch[1];
}
}
}

Node* Find(int x) {
Node *u = root,*ret = R;int rr = inf;
while(u != null) {
if(u -> val == x)return u;
if(u -> val < x)u = u -> ch[1];
else {
if(u -> val < rr) {
rr = u -> val;ret = u;
}
u = u -> ch[0];
}
}
return ret;
}

void Delete(int x) {
Node *now = Find(x);
Splay(now,null);Splay(L,root);
ans += root -> ch[0] -> ch[1] -> sz;
root -> ch[0] -> ch[1] = null;
Splay(root -> ch[0],null);
}

int query(int x) {
if(x > root -> sz-2)return -1;
int r = root -> sz - x;
Node *u = root;
while(u != null) {
int ret1 = u -> ch[0] -> sz,ret2 = ret1 + u -> num;
if(r > ret1 && r <= ret2)return u -> val + k;
if(r <= ret1)u = u -> ch[0];
else {
r -= ret2;
u = u -> ch[1];
}
}
}

char s[5];

int main() {
int x;init();
scanf("%d%d",&n,&minn);
for(int i = 1; i <= n; i++) {
scanf("%s%d",s,&x);
if(s[0] == 'I') {
if(x >= minn)Insert(x-k);
}
else if(s[0] == 'A') {
k += x;
}
else if(s[0] == 'S'){
k -= x;
Delete(minn-k);//把小于min-k的都删去
}
else if(s[0] == 'F'){
printf("%d\n",query(x));
}
}
printf("%d\n",ans);
return 0;
}