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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;}

 

HDU 5884 Sort (二分+k叉哈夫曼树)