能把 not working now 写成 not working hard now
还查一晚上也是没谁了
我的做法是维护两个set 分别是前20% 和后80%
#include<iostream>
#include<algorithm>
#include<set>
#include<cmath>
#include<map>
#include<cstdio>
using namespace std;
#define sz(X) ((int)X.size())
int N,M;
char nam[70005][25];
map<string, int> mp;
struct Node{
int id; int val; int ti;
Node(int a=0, int b=0, int c=0):id(a),val(b),ti(c){}
bool operator < (const Node &v) const{
if(val != v.val) return val > v.val;
return ti > v.ti;
}
bool operator == (Node v) const{
return id==v.id && val==v.val && ti==v.ti;
}
}mem[70005];
set<Node> s1; set<Node> s2;
set<Node>::iterator it1,it,it2;
int main(){
while(~scanf("%d",&N)) {
mp.clear();
s1.clear(); s2.clear();
for(int i = 0; i < N; ++i) {
int a; scanf("%s %d",nam[i], &a);
s2.insert(Node(i,a,i));
mem[i] = Node(i,a,i);
mp[nam[i]] = i;
}
int tot = N;
int K = floor(N/5);
while(K--) {
it = s2.begin();
Node y = *it;
s2.erase(y);
s1.insert(y);
}
scanf("%d",&M);
for(int i = N; i < N+M; ++i) {
char s[5]; scanf("%s",s);
if(s[0] == '+') {
int a; scanf("%s %d",nam[i],&a);
Node tt = Node(i,a,i);
mem[i] = tt; mp[nam[i]] = i;
tot ++;
int x = floor(tot/5); int y = tot-x;
if(sz(s2) == y-1) {
s1.insert(tt);
it = --s1.end(); Node t2 = *it; s1.erase(it);
if(t2 == tt) {
printf("%s is not working now.\n",nam[i]);
s2.insert(t2);
}else {
printf("%s is working hard now.\n",nam[i]);
printf("%s is not working now.\n",nam[t2.id]);
s2.insert(t2);
}
}else {
s2.insert(tt);
it = s2.begin(); Node t2 = *it; s2.erase(it);
if(t2 == tt) {
printf("%s is working hard now.\n",nam[i]);
s1.insert(t2);
}else {
printf("%s is not working now.\n",nam[i]);
printf("%s is working hard now.\n",nam[t2.id]);
s1.insert(t2);
}
}
}else if(s[0] == '-') {
char _s[25];
scanf("%s",_s);
Node tt = mem[mp[_s]];
it1 = s1.find(tt); it2 = s2.find(tt);
if(it1 != s1.end()) s1.erase(it1);
else s2.erase(it2);
tot--;
int x = floor(tot/5); int y = tot-x;
if(sz(s1) > x) {
it = --s1.end(); Node t2 = *it;
printf("%s is not working now.\n",nam[t2.id]);
s1.erase(it); s2.insert(t2);
}
if(sz(s2) > y) {
it = s2.begin(); Node t2 = *it;
printf("%s is working hard now.\n",nam[t2.id]);
s2.erase(it); s1.insert(t2);
}
}
}
}
return 0;
}