HDU - 5884

一年前入门的题目了，当时没写出来，现在填坑

题目

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 than k 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 than T cost. So Bob wants to know the smallest k to make the program complete in time.

坑点

1. 明显用二分 + 贪心，二分k,每次选择cost最少的来合并

2. 假如枚举的 k，那么每次合并消耗的是 k−1个，最后合并完后剩下一个，所以总消耗是 n−1 个，如果 (n−1)%（k-1）!=0，会存在零头，要先合并 零头 + 1个序列

3. 有点卡常

方法1

1. 用priority_queue有点卡，可以先把n个序列从小到大排序，优先队列每次要check函数内定义，n <= 10^5，这样优先队列内存占用小于 1M （栈空间） 所以可以开的下，输入输出用scanf printf

2. 时间复杂度 O(n * logn * logn j) 出队入队次数 * 二分次数 * 优先队列单次插入删除

3. code

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N = 1e5 + 10;
int a[N];

ll n,t;    

bool check(int k){
    
    priority_queue< int, vector<int> , greater<int> > q1;
    for(int i = 0; i < n; ++i)
        q1.push(a[i]);

    int ct = 2,sz = (n - 1) %  (k - 1);
    ll res = 0;
    if(sz){
        for(int i = 0; i <= sz; ++i){
            res += q1.top();
            q1.pop();
        }
        q1.push(res);
    }

    ll z;
    while(q1.size() > 1 && res <= t){
        z = 0;
        for(int i = 0; i < k && q1.size(); ++i){
            z += q1.top();
            q1.pop();
        }
        q1.push(z);
        res += z;
    }
    
    return res <= t;
} 

int main(){
    
    int T; scanf("%d",&T);
    while(T--){

        scanf("%lld %lld",&n,&t);

        
        for(int i = 0; i < n; ++i)
        {
            scanf("%d",&a[i]);        
        }
        sort(a, a + n);
        
        int l = 2,r = n,ans = n;
        while(l <= r && l < ans){
            int m = (l + r) >> 1;
            if(check(m)){
                ans = min(ans,m);
                r = m - 1;
            }else l = m + 1;
        }
        printf("%d\n",ans);
    }
    return 0;
}

方法2

1. 先排序，然后用两个队列模拟优先队列,这样就少了一个log。很好过

2. 时间复杂度 O（nlogn)

3. code

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N = 1e5 + 10;
int a[N];

ll n,t;    

bool check(int k){
    
    queue<ll> q1,q2;
    for(int i = 0; i < n; ++i)
        q1.push(a[i]);

    int ct = 2,sz = (n - 1) %  (k - 1);
    ll res = 0;
    if(sz){
        for(int i = 0; i <= sz; ++i){
            res += q1.front();
            q1.pop();
        }
        q1.push(res);
    }

    ll x,y;
    while(q1.size() + q2.size() > 1 && res <= t){
        ll z = 0;
        for(int i = 0; i < k && q1.size() + q2.size(); ++i){
            if(q1.size() && q2.size()){
                x = q1.front();
                y = q2.front();
                if(x < y){
                    z += x;
                    q1.pop();
                }else{
                    z += y;
                    q2.pop();
                }
            }else if(q1.size()){
                ct = 2;
                z += q1.front();
                q1.pop();
            }else{
                ct = 1;
                z += q2.front();
                q2.pop();
            }
        }
        if(ct == 1)
            q1.push(z);
        else q2.push(z);
        res += z;
    }
    
    return res <= t;
} 

int main(){
    
    int T; scanf("%d",&T);
    while(T--){

        scanf("%lld %lld",&n,&t);

        
        for(int i = 0; i < n; ++i)
        {
            scanf("%d",&a[i]);        
        }
        sort(a, a + n);
        
        int l = 2,r = n,ans = n;
        while(l <= r && l < ans){
            int m = (l + r) >> 1;
            if(check(m)){
                ans = min(ans,m);
                r = m - 1;
            }else l = m + 1;
        }
        printf("%d\n",ans);
    }
    return 0;
}