题意:中文题。
析:用treap 维护一个序列,用一个变量来记录全体人员加的工资和扣除的工资,对于每次扣除时,把一小于的最低工资的删除,很容易维护。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
//#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 1000;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
int val, ans;
struct Node{
Node *ch[2];
int r, v, s;
Node(int vv) : v(vv) { ch[0] = ch[1] = 0; r = rand(); s = 1; }
bool operator < (const Node &rhs) const{
return r < rhs.r;
}
int cmp(int x) const{
if(x == v) return -1;
return x < v ? 0 : 1;
}
void maintain(){
s = 1;
if(ch[0]) s += ch[0]->s;
if(ch[1]) s += ch[1]->s;
}
};
void rotate(Node *&o, int d){
Node *k = o->ch[d^1];
o->ch[d^1] = k->ch[d];
k->ch[d] = o;
o->maintain();
k->maintain(); o = k;
}
void insert(Node *&o, int x){
if(o == 0) o = new Node(x);
else{
int d = (x < o->v ? 0 : 1);
insert(o->ch[d], x);
if(o->ch[d]->r > o->r) rotate(o, d^1);
}
o->maintain();
}
void del(Node *&o){
if(o == 0) return ;
++ans;
del(o->ch[0]);
del(o->ch[1]);
delete o;
o = 0;
}
void remove(Node *&o, int x){
if(o == 0) return ;
while(o && x > o->v){
del(o->ch[0]);
if(o->ch[1]) rotate(o, 0);
else del(o);
}
if(o == 0) return ;
remove(o->ch[0], x);
o->maintain();
}
int get_kth(Node *o, int k){
if(o == 0 || k <= 0 || k > o->s) return -1;
int s = o->ch[1] == 0 ? 0 : o->ch[1]->s;
if(k == s + 1) return o->v + val;
else if(k <= s) return get_kth(o->ch[1], k);
return get_kth(o->ch[0], k-s-1);
}
Node *root;
int main(){
srand(233333);
scanf("%d %d", &n, &m);
int x; root = 0;
char op[5];
while(n--){
scanf("%s %d", op, &x);
if(op[0] == ‘I‘){
if(x < m) continue;
insert(root, x - val);
}
else if(op[0] == ‘A‘) val += x;
else if(op[0] == ‘S‘){
val -= x; x = m - val;
remove(root, x);
}
else printf("%d\n", get_kth(root, x));
}
printf("%d\n", ans);
return 0;
}
原文:http://www.cnblogs.com/dwtfukgv/p/7624425.html