HDU 5884 k叉的哈夫曼 O（n）构造

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Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1832    Accepted Submission(s): 458

Problem Description

Recently, Bob has just learnt a naive sorting algorithm: merge sort. Now, Bob receives a task from Alice.
Alice will give Bob N sorted sequences, and the i-th sequence includes ai elements. Bob need to merge all of these sequences. He can write a program, which can merge no more thank sequences in one time. The cost of a merging operation is the sum of the length of these sequences. Unfortunately, Alice allows this program to use no more thanT cost. So Bob wants to know the smallest k to make the program complete in time.

 

Input

The first line of input contains an integer t0, the number of test cases. t0 test cases follow.
For each test case, the first line consists two integers N (2≤N≤100000) and T (∑Ni=1ai<T<231).
In the next line there are N integers a1,a2,a3,...,aN(∀i,0≤ai≤1000).

 

Output

For each test cases, output the smallest k.

 

Sample Input

15 251 2 3 4 5

 

Sample Output

3

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

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#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <ctime>#include <vector>#include <cstdio>#include <cctype>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf (-((LL)1<<40))#define lson k<<1, L, (L + R)>>1#define rson k<<1|1,  ((L + R)>>1) + 1, R#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define mem(a, b) memset(a, b, sizeof(a))#define FIN freopen("in.txt", "r", stdin)#define FOUT freopen("out.txt", "w", stdout)#define rep(i, a, b) for(int i = a; i <= b; i ++)#define dec(i, a, b) for(int i = a; i >= b; i --)typedef long long ll;ll a[100005];int n,m;int work(int x){    queue<ll>q,d;    int yy=(n-1)%(x-1);    if(yy!=0)    {        for(int i=0; i<x-1-yy; i++)            q.push(0);    }    for(int i=1; i<=n; i++)        q.push(a[i]);    ll ans=0;    while(!q.empty()|| !d.empty())    {        ll sum=0;        for(int i=0; i<x; i++)        {            if(q.empty() &&d.empty())                break;            if(q.empty())            {                sum+=d.front();                d.pop();            }            else if(d.empty())            {                sum+=q.front();                q.pop();            }            else            {                int u=q.front();                int v=d.front();                if(u<v)                {                    sum+=u;                    q.pop();                }                else                {                    sum+=v;                    d.pop();                }            }        }        ans+=sum;        if(q.empty()&&d.empty())            break;        d.push(sum);    }    if(ans>m)        return 0;    return 1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1; i<=n; i++)            scanf("%lld",&a[i]);        sort(a+1,a+1+n);        int l=2,r=n;        while(l<r)        {            int mid=(l+r)>>1;            if(!work(mid))                l=mid+1;            else                r=mid;        }        printf("%d\n",l);    }    return 0;}