【暑假】[实用数据结构]UVa11997 K Smallest Sums

UVa11997 K Smallest Sums

 题目：

K Smallest Sums

You‘re given k arrays, each array has k integers. There are k^k 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

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思路：

 先思考2个有序表（sort后）AB合成的情况。组织如下

【 表】1： A₁+B₁ <=  A₁+B₂ <=  A₁+B₃ <=......

【表】2：A₂+B₁ <=  A₂+B₂ <=  A₂+B₃ <=......

......

【表】n：A_n+B₁ <=  A_n+B₂ <=  A_n+B₃ <=......

转化为多路归并问题且有总数限制为k，因此将n个【表】合并为一个有序【表】C且表中数据数目为k。优先队列处理：初始化优先队列为n个表的第一元素，每次在队列中取出最小元素加入C并将该最小元素在【表】中的后一个元素入队。共取k个元素。

对于k个有序表的情况：两两合并。

于是有代码如下：

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<queue>
 4 #define FOR(a,b,c) for(int a=(b);a<(c);a++)
 5 using namespace std;
 6
 7 const int maxn = 800 + 10;
 8
 9 struct Node{
10     int s,b;
11     bool operator <(const Node& rhs) const{ //相反定义 <
12       return s>rhs.s;
13     }
14 };
15
16
17 void merge(int* A,int* B,int*C,int n){ //合并AB数组取其前n小的和放入C
18 priority_queue<Node> Q;
19   FOR(i,0,n) Q.push(Node{A[i]+B[0],0});
20   FOR(i,0,n){
21       Node u=Q.top();Q.pop();
22       C[i]=u.s;
23       int b=u.b;
24       if(b+1 < n) Q.push(Node{u.s-B[b]+B[b+1],b+1}); //加入A[a][b+1]
25   }
26 }
27
28 int main(){
29 int n;
30 int A[maxn],B[maxn];
31
32   while(scanf("%d",&n)==1){
33       FOR(j,0,n) scanf("%d",&A[j]); sort(A,A+n);
34       FOR(i,1,n){
35            FOR(j,0,n) scanf("%d",&B[j]);
36            sort(B,B+n);
37          merge(A,B,A,n);
38     }
39     printf("%d",A[0]);
40     FOR(i,1,n)
41       printf(" %d",A[i]);
42     printf("\n");
43  }
44  return 0;
45 }