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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。
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
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原文地址:http://www.cnblogs.com/zjc-Connor/p/5953808.html