HDU 5884 Sort (二分+k叉哈夫曼树)

题意：n 个有序序列的归并排序.每次可以选择不超过 k 个序列进行合并,合并代价为这些序列的长度和.总的合并代价不能超过T, 问 k最小是多少。

析：首先二分一下这个 k 。然后在给定 k 的情况下，这个代价其实就是 k 叉的哈夫曼树问题。然后用两个队列维护一下就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
LL K;

bool judge(int k){
    int x = (n-1) % (k-1);
    queue<LL> q, d;
    for(int i = 0; i < k-x-1; ++i) q.push(0);
    for(int i = 0; i < n; ++i)  q.push(a[i]);
    LL ans = 0;
    while(!q.empty() || !d.empty()){
        LL tmp = 0;
        for(int i = 0; i < k; ++i){
            if(!q.empty() && !d.empty()){
                LL s = q.front();
                LL t = d.front();
                if(s < t){  q.pop();  tmp += s;  }
                else{  d.pop();  tmp += t; }
            }
            else if(!q.empty()){
                tmp += q.front();
                q.pop();
            }
            else if(!d.empty()){
                tmp += d.front();
                d.pop();
            }
            else  break;
        }

        ans += tmp;
        if(q.empty() && d.empty())  break;
        d.push(tmp);
    }
    return ans <= K;
}

int solve(){
    int l = 2, r = n;
    while(l < r){
        int mid = (l+r)>>1;
        if(judge(mid))  r = mid;
        else l = mid + 1;
    }
    return l;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d %I64d", &n, &K);
        for(int i = 0; i < n; ++i)  scanf("%d", a+i);
        sort(a, a+n);
        int ans = solve();
        printf("%d\n", ans);
    }
    return 0;
}

 

来源：https://www.cnblogs.com/dwtfukgv/p/5880063.html