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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)
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