题解 P1168 【中位数】
这题可以用两个堆做。。。左堆为大根堆右堆,为小根堆,遇到大于大根堆堆顶的就放在小根堆里。。。反之放入大根堆。。让他们的元素个数差值为一,如果不为一就从多的一个拿出来放入少的一个中。。。之后元素多的堆顶的元素就是中位数了。。哈哈。。下面是代码
#include<cstdio>
#include<iostream>
#include<queue>
#include<cmath>
#include<algorithm>
#include<ctime>
using namespace std;
inline int read()
{
char ch=getchar();
int a=0,t=1;
while(ch<'0'||ch>'9') {if(ch=='-') t=-1;ch=getchar();}
while(ch<='9'&&ch>='0') {a=a*10+ch-'0';ch=getchar();}
return a*t;
}
priority_queue<int,vector<int> > l;
priority_queue<int,vector<int>,greater<int> > r;
int n;
int num=0;
int leftnum=0,rightnum=0;
int x,y;
int a[5000001];
int main()
{
//int th=clock();
freopen("wq.in","r",stdin);freopen("wq.out","w",stdout);
n=read();
//l.resize(1000001);r.resize(1000001);
a[1]=read();a[2]=read();a[3]=read();
printf("%d\n",a[1]);
sort(a+1,a+4);
printf("%d\n",a[2]);
l.push(a[1]);leftnum++;
l.push(a[2]);leftnum++;
r.push(a[3]);rightnum++;
for(int q=5;q<=n;q+=2)
{
for(int i=q-1;i<=q;++i)
{
a[i]=read();
//if(l.empty()==1) {l.push(a[i]);leftnum++;continue;}
//if(r.empty()==1) {r.push(a[i]);rightnum++;continue;}
x=l.top();y=r.top();
int bo=0;
if(a[i]<=x) {l.push(a[i]),leftnum++,bo=1;continue;}
if(a[i]>=y) {r.push(a[i]),rightnum++,bo=1;continue;}
if(a[i]>x) {r.push(a[i]),rightnum++,bo=1;continue;}
if(a[i]<y) {l.push(a[i]),leftnum++,bo=1;continue;}
//if(bo==0) l.push(a[i]),leftnum++;
}
while(abs(leftnum-rightnum)!=1)
{
if(leftnum>rightnum)
{
x=l.top();
l.pop();
leftnum--;
r.push(x);
rightnum++;
continue;
}
if(leftnum<rightnum)
{
x=r.top();
r.pop();
rightnum--;
l.push(x);
leftnum++;
continue;
}
}
if(leftnum>rightnum)
{
x=l.top();
//num++;
printf("%d\n",x);
continue;
}
if(leftnum<rightnum)
{
//num++;
x=r.top();
printf("%d\n",x);
continue;
}
}
//while(l.empty()!=0)
//printf("%d",th-clock());
//printf("%d",num);
return 0;
}