ural1316 Electronic Auction
Electronic Auction
Memory limit: 64 MB
Input
- “BID X” — a customer announces that he is ready to buy a pig at a price of X bibriks;
- “DEL X” — a customer cancels his bid for a pig at a price of X bibriks;
- “SALE X K” — a seller puts up for sale К pigs at a price of X bibriks.
X is between 0.01 and 10000.00 bibriks and has at most 2 digits after the decimal point. K is an integer between 1 and 100000. The number of operations does not exceed 100000. All operations are correct. The last line contains the word “QUIT”.
Output
Sample
input | output |
---|---|
BID 0.01 |
0.06 |
分析:离散化+树状数组;
代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <hash_map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=1e5+;
const int dis[][]={,,-,,,-,,};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p%mod;p=p*p%mod;q>>=;}return f;}
int n,m,k,t,a[maxn],num;
double c[maxn];
ll ans;
struct node
{
char a[];
double b;
int c;
}op[maxn];
void add(int x,int y)
{
for(int i=x;i<=num;i+=(i&(-i)))
{
a[i]+=y;
}
}
int get(int x)
{
int ans=;
for(int i=x;i;i-=(i&(-i)))
ans+=a[i];
return ans;
}
int main()
{
int i,j;
while(~scanf("%s",op[++n].a)&&op[n].a[]!='Q')
{
scanf("%lf",&op[n].b);
c[n]=op[n].b;
if(op[n].a[]=='S')scanf("%d",&op[n].c);
}
n--;
sort(c+,c+n+);
num=unique(c+,c+n+)-c-;
rep(i,,n)op[i].b=lower_bound(c+,c+num+,op[i].b)-c;
rep(i,,n)
{
if(op[i].a[]=='B')add((int)op[i].b,);
else if(op[i].a[]=='D')add((int)op[i].b,-);
else ans+=min(op[i].c,get(num)-get((int)op[i].b-));
}
printf("%.2f\n",(double)ans/);
//system("Pause");
return ;
}
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