HDU 5884 Sort (二分)

Posted 2020-08-09 Recoder

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5884

nn个有序序列的归并排序.每次可以选择不超过kk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TT, 问kk最小是多少

用一个队列维护合并的数，二分一下判断合理性。注意一点的是要是(n - 1)%(k - 1) != 0的话，就要先合并前(n - 1)%(k - 1) + 1项，这样会更优一点。还有细节问题很多要注意。

 1 //#pragma comment(linker, "/STACK:102400000, 102400000")
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <vector>
 8 #include <cmath>
 9 #include <ctime>
10 #include <list>
11 #include <set>
12 #include <map>
13 #include <queue>
14 using namespace std;
15 typedef long long LL;
16 typedef pair <int, int> P;
17 const int N = 1e5 + 5;
18 LL a[N], m;
19 int n;
20 
21 bool judge(int k) {
22     queue <int> que;
23     while(!que.empty()) {
24         que.pop();
25     }
26     int sum = 0, i = 1;
27     if((n - 1)%(k - 1)) {
28         for( ; i <= (n - 1) % (k - 1) + 1; ++i) {
29             sum += a[i];
30         }
31         que.push(sum);
32     }
33     int ans = sum;
34     for(; i <= n || !que.empty(); ++i) {
35         if(que.size() <= 1 && i > n)
36             break;
37         int cnt = k, ok = 0;
38         sum = 0;
39         while(cnt--) {
40             if((que.size() && que.front() < a[i])|| i > n) {
41                 sum += que.front();
42                 que.pop();
43             } else {
44                 ok = 1;
45                 sum += a[i++];
46             }
47         }
48         if(ok) {
49             --i;
50         }
51         ans += sum;
52         que.push(sum);
53     }
54     return ans <= m;
55 }
56 
57 int main()
58 {
59     int t;
60     scanf("%d", &t);
61     while(t--) {
62         scanf("%d %lld", &n, &m);
63         for(int i = 1; i <= n; ++i) {
64             scanf("%lld", a + i);
65         }
66         sort(a + 1, a + n + 1);
67         int l = 1, r = n;
68         while(l < r) {
69             int mid = (l + r) / 2;
70             if(judge(mid)) {
71                 r = mid;
72             } else {
73                 l = mid + 1;
74             }
75         }
76         printf("%d\n", r);
77     }
78     return 0;
79 }