Problem K
K Smallest Sums
You're given k arrays, each array has k integers. There are kk ways to pick exactly one element in each array and calculate the sum of the integers. Your task is to find the k smallest sums among them.
Input
There will be several test cases. The first line of each case contains an integer k (2<=k<=750). Each of the following k lines contains k positive integers in each array. Each of these integers does not exceed 1,000,000. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.
Output
For each test case, print the k smallest sums, in ascending order.
Sample Input
3
1 8 5
9 2 5
10 7 6
2
1 1
1 2
Output for the Sample Input
9 10 12
2 2
Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <string> 7 #include <vector> 8 #include <stack> 9 #include <queue> 10 #include <set> 11 #include <map> 12 #include <list> 13 #include <iomanip> 14 #include <cstdlib> 15 #include <sstream> 16 using namespace std; 17 typedef long long LL; 18 const int INF=0x5fffffff; 19 const double EXP=1e-6; 20 const int MS=800; 21 int ans[MS],a[MS],n; 22 struct node 23 { 24 int s,b; 25 node(int s,int b):s(s),b(b){} 26 bool operator <(const node &a)const 27 { 28 return s>a.s; 29 } 30 }; 31 32 void merge(int *A,int *B,int *C,int n) 33 { 34 priority_queue<node> pq; 35 for(int i=0;i<n;i++) 36 pq.push(node(A[i]+B[0],0)); 37 for(int i=0;i<n;i++) 38 { 39 node t=pq.top(); 40 pq.pop(); 41 C[i]=t.s; 42 int b=t.b; 43 if(b+1<n) 44 pq.push(node(t.s-B[b]+B[b+1],b+1)); 45 } 46 } 47 48 int main() 49 { 50 while(scanf("%d",&n)!=EOF) 51 { 52 for(int i=0;i<n;i++) 53 scanf("%d",&ans[i]); 54 sort(ans,ans+n); 55 for(int i=1;i<n;i++) 56 { 57 for(int j=0;j<n;j++) 58 scanf("%d",&a[j]); 59 sort(a,a+n); 60 merge(ans,a,ans,n); 61 } 62 for(int i=0;i<n;i++) 63 { 64 if(i) 65 printf(" "); 66 printf("%d",ans[i]); 67 } 68 printf(" "); 69 } 70 return 0; 71 }