Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2522 Accepted Submission(s): 1138
Special Judge

Problem Description

In ACM/ICPC contest, the ''Dirt Ratio'' of a team is calculated in the following way. First let's ignore all the problems the team didn't pass, assume the team passed

Picture from MyICPC

Input

The first line of the input contains an integer

Output

For each test case, print a single line containing a floating number, denoting the lowest ''Dirt Ratio''. The answer must be printed with an absolute error not greater than

Sample Input

1 5 1 2 1 2 3

Sample Output

0.5000000000

Hint

For every problem, you can assume its final submission is accepted.

Source

2017 Multi-University Training Contest - Team 4

求一个序列中的连续子序列S，使得（S中不同元素的个数）/（S的长度）最小化。挺不错的一道题，很考验思维。对于这种最小化的题目很容易往二分上面想，得到 a/b<=k ,我们只要不断二分k，找到下界即可，问题是对于k如何判定是否可行，枚举所有区间显然不靠谱，考虑这个式子 diff(l,r)/(r-l+1)<=k -> diff(l,r)+k*l<=k*(r+1), 我们可以枚举一下右端点，然后找到一个最优的左端点判断能否使得这个不等式成立即可，可以用线段树区间修改来维护这个值，结点 u(L,R),保存的就是[L,R]中最小的 diff(l,r)+k*l ,显然l€[L,R]。初始化根节点为k*L,然后遍历右端点的时候更新对应的区间，也就是会对那些区间的diff造成变化。

  1 #include<iostream>
  2 #include<cstring>
  3 #include<queue>
  4 #include<cstdio>
  5 #include<stack>
  6 #include<set>
  7 #include<map>
  8 #include<cmath>
  9 #include<ctime>
 10 #include<time.h> 
 11 #include<algorithm>
 12 #include<bits/stdc++.h>
 13 using namespace std;
 14 #define mp make_pair
 15 #define pb push_back
 16 #define debug puts("debug")
 17 #define LL long long 
 18 #define pii pair<int,int>
 19 #define inf 0x3f3f3f3f
 20 
 21 #define mid ((L+R)>>1)
 22 #define lc (id<<1)
 23 #define rc (id<<1|1)
 24 #define eps 1e-5
 25 const int _M=60006;
 26 double sum[_M<<2];
 27 int laz[_M<<2];
 28 int a[_M],x[_M],pre[_M];
 29 void pushdown(int id){
 30     if(laz[id]){
 31         laz[lc]+=laz[id];
 32         laz[rc]+=laz[id];
 33         sum[lc]+=laz[id];
 34         sum[rc]+=laz[id];
 35         laz[id]=0;
 36     }
 37 }
 38 void pushup(int id){
 39     sum[id]=min(sum[lc],sum[rc]);
 40 }
 41 void build(int id,int L,int R,double k){
 42     if(L==R){
 43         sum[id]=k*L;
 44         return;
 45     }
 46     build(lc,L,mid,k);
 47     build(rc,mid+1,R,k);
 48     pushup(id);
 49 }
 50 void update(int id,int L,int R,int l,int r){
 51     if(L>=l&&R<=r){
 52         sum[id]++;
 53         laz[id]++;
 54         return ;
 55     }
 56     pushdown(id);
 57     if(l<=mid) update(lc,L,mid,l,r);
 58     if(r>mid) update(rc,mid+1,R,l,r);
 59     pushup(id);
 60 }
 61 double query(int id,int L,int R,int l,int r){
 62     if(L>=l&&R<=r){
 63         return sum[id];
 64     }
 65     pushdown(id);
 66     if(r<=mid) return query(lc,L,mid,l,r);
 67     else if(l>mid) return query(rc,mid+1,R,l,r);
 68     else return min(query(lc,L,mid,l,r),query(rc,mid+1,R,l,r));
 69     pushup(id);
 70 }
 71 bool ok(double k,int n){
 72     memset(sum,0,sizeof(sum));
 73     memset(laz,0,sizeof(laz));
 74     build(1,1,n,k);
 75     for(int i=1;i<=n;++i){
 76         update(1,1,n,pre[i]+1,i);
 77         if(query(1,1,n,1,i)<=k*(i+1)) return 1;
 78     }
 79     return 0;
 80 }
 81 int main(){
 82     int t,n,m,i,j;
 83     scanf("%d",&t);
 84     while(t--){
 85         scanf("%d",&n);
 86         memset(x,0,sizeof(x));
 87         for(i=1;i<=n;++i) {
 88             scanf("%d",a+i);
 89             pre[i]=x[a[i]];
 90             x[a[i]]=i;
 91         }
 92         
 93         double l=0,r=1.0;
 94         while(abs(l-r)>eps){
 95             double k=(l+r)/2;
 96             if(ok(k,n)) r=k;
 97             else l=k+eps;
 98         }
 99         printf("%.10f\n",l);
100     }
101     return 0;
102 }

HDU-6070-二分+线段树

Dirt Ratio

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