hdu 6070 Dirt Ratio（二分+线段树）（2017 Multi-University Training Contest - Team 4 ）

Dirt Ratio

题目链接：Dirt Ratio

题意：x为区间数字的种数，y为区间长度，求x/y的最小值。

官方题解：



size(l,r)r−l+1<=mid转化为size(l,r)+mid∗l<=mid∗(r+1)是关键性的一步。

转化后我们就可以通过线段树来维护size(l,r)+mid∗l的最小值了

线段树保存的是size(l,r)+mid∗l，然后枚举右端点r，判断如果在区间[i，r]（1<=i<=r）满足上式，那就可以缩小范围继续枚举答案

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int maxn=6e4+10;
const double inf=9999999.0;
const double eps=1e-5;
struct Segtree
{
double val,lazy;
int le,ri;
int mid()
{
return (le+ri)>>1;
}
} tree[maxn<<2];
int last_apper[maxn],a[maxn];
int n;

void Build(int rt,int le,int ri,double num)
{
tree[rt].lazy=0;//注意lazy不要忘记清0
tree[rt].le=le,tree[rt].ri=ri;
if(le==ri)
{
tree[rt].val=num*le;
return ;
}
int mid=tree[rt].mid();
Build(rt<<1,le,mid,num);
Build(rt<<1|1,mid+1,ri,num);
tree[rt].val=min(tree[rt<<1].val,tree[rt<<1|1].val);
}

void Pushdown(int rt)
{
tree[rt<<1].lazy+=tree[rt].lazy;
tree[rt<<1].val+=tree[rt].lazy;
tree[rt<<1|1].lazy+=tree[rt].lazy;
tree[rt<<1|1].val+=tree[rt].lazy;
tree[rt].lazy=0;
}

void Update(int rt,int le,int ri,int num)
{
if(le<=tree[rt].le&&tree[rt].ri<=ri)
{
tree[rt].lazy+=num;
tree[rt].val+=num;
return ;
}
if(tree[rt].lazy)
Pushdown(rt);
int mid=tree[rt].mid();
if(le<=mid)
Update(rt<<1,le,ri,num);
if(ri>mid)
Update(rt<<1|1,le,ri,num);
tree[rt].val=min(tree[rt<<1].val,tree[rt<<1|1].val);
}

double Query(int rt,int le,int ri)
{
if(le<=tree[rt].le&&tree[rt].ri<=ri)
return tree[rt].val;
if(tree[rt].lazy)
Pushdown(rt);
int mid=tree[rt].mid();
double minn=inf;
if(le<=mid)
minn=Query(rt<<1,le,ri);
if(ri>mid)
minn=min(minn,Query(rt<<1|1,le,ri));
tree[rt].val=min(tree[rt<<1].val,tree[rt<<1|1].val);
return minn;
}

bool judge(double x)
{
Build(1,1,n,x);
memset(last_apper,0,sizeof(last_apper));
for(int i=1; i<=n; ++i)
{
Update(1,last_apper[a[i]]+1,i,1);
last_apper[a[i]]=i;
if(Query(1,1,i)<=x*(i+1))
return true;
}
return false;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1; i<=n; ++i)
scanf("%d",&a[i]);
double l=0,r=1,ans=0;
while(l<=r)
{
double mid=(l+r)/2.0;
if(judge(mid))
{
ans=mid;
r=mid-eps;
}
else l=mid+eps;
}
printf("%.10lf\n",ans);
}
return 0;
}