UVA.11997- K Smallest Sums, OJ4TH.368 - Magry's Sum I

感谢算法助教们给了这么一个好题...

题意：

给出n个数组，每个数组有n个元素，我们从每个数组中挑选一个元素，共计n个元素求和，得到共计 $ k^k $ 种sum，求sum中的最小n个值。

1. 利用二分查找符合条件的最大sum值，dfs搜索判断二分条件。最坏时间复杂度 $O(n^{2}log(n*Max(a[i][j])) $
2. 刘汝佳训练指南上给出了优先队列多路归并的思想。

思路1代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 using namespace std;
 6
 7 int a[751][751];
 8 int pos[751];
 9 int n;
10 int mid;
11 int sum;
12 int ans[1000];
13 int cnt = 0;
14 void dfs(int i)
15 {
16     if (sum > mid) return;
17     ans[cnt ++] = sum;
18     if (cnt > n) {
19         return;
20     }
21     for (i; i < n && cnt < n; i ++)
22     {
23         if (pos[i] == n - 1) continue;
24
25         ++pos[i];
26         sum += a[i][ pos[i] ] - a[i][ pos[i] - 1 ];
27         dfs(i);
28         sum -= a[i][ pos[i] ] - a[i][ pos[i] - 1 ];
29         --pos[i];
30     }
31 }
32
33 int main()
34 {
35     while(~scanf("%d",&n))
36     {
37         for (int i = 0; i < n; i++)
38         for (int j = 0; j < n; j++)
39             scanf("%d",&a[i][j]);
40
41         for (int i = 0; i < n; i++)
42             sort(a[i],a[i] + n);
43
44         memset(pos,0,sizeof pos);
45         int l = -1, r = n*1000000;
46         while(r-l > 1)
47         {
48             sum = 0;
49             for (int i = 0; i < n; i++)
50                 sum += a[i][0];
51
52             cnt = 0;
53             mid = (l+r) / 2;
54             dfs(0);
55             if (cnt >= n) r = mid;
56             else l = mid;
57         }
58         cnt = 0;
59         sum = 0;
60         for (int i = 0; i < n; i++)
61                 sum += a[i][0];
62
63         mid = r - 1;
64         dfs(0);
65         sort(ans,ans+cnt);
66         for (int i = 0; i < cnt; i++)
67             printf("%d ",ans[i]);
68         for (int i = cnt; i < n-1; i++)
69             printf("%d ",r);
70         printf("%d\n",r);
71     }
72 }

思路2代码留坑

UVA.11997- K Smallest Sums, OJ4TH.368 - Magry's Sum I