K个最小和（K Smallest Sums）

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

分析：本题也是用优先级队列来解决多路归并问题。本题输入数据有多行，每次处理两行，然后从第一列开始处理，之后两两合并。

初步代码如下：

#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 768;
int A[maxn][maxn];
struct Node 
{
	int sum,b;
	Node(int s,int b) : sum(s) , b(b) {}
	bool operator<(const Node & obj) const
	{
		return sum > obj.sum;
	}
};
void merge(int *A,int *B,int *C,int n)
{
	priority_queue<Node> pq;
	for (int i = 0; i < n; i++)
		pq.push(Node(A[i] + B[0],0));
	for (int j = 0; j < n; j++)
	{
		Node item = pq.top();
		pq.pop();
		C[j] = item.sum;
		int b = item.b;
		if (b + 1 < n)
			pq.push(Node(item.sum - B[b] + B[b + 1], b + 1));//取下一个数
	}
}
int main()
{
	int n;
	while (scanf("%d",&n) == 1)
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
				scanf("%d",&A[i][j]);
			sort(A[i],A[i] + n);
		}
		for (int k = 1; k < n; k++)//两两合并
			merge(A[0],A[k],A[0],n);
		printf("%d",A[0][0]);
		for (int j = 1; j < n; j++)
			printf(" %d",A[0][j]);
		printf("\n");
	}
	return 0;
}

仔细考虑下，会发现以上代码可以改进，其实一开始并不需要把K*K元素都保存。可以每次只读k个元素，这样大大降低了空间复杂度，改进后的代码如下：

#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 768;
int A[maxn],B[maxn];
struct Node 
{
	int sum,b;
	Node(int s,int b) : sum(s) , b(b) {}
	bool operator<(const Node & obj) const
	{
		return sum > obj.sum;
	}
};
void merge(int *A,int *B,int *C,int n)
{
	priority_queue<Node> pq;
	for (int i = 0; i < n; i++)
		pq.push(Node(A[i] + B[0],0));
	for (int j = 0; j < n; j++)
	{
		Node item = pq.top();
		pq.pop();
		C[j] = item.sum;
		int b = item.b;
		if (b + 1 < n)
			pq.push(Node(item.sum - B[b] + B[b + 1], b + 1));
	}
}
int main()
{
	int n;
	while (scanf("%d",&n) == 1)
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)//每次读一行
				scanf("%d",&A[j]);
			sort(A,A + n);
			if (i != 0)
				merge(B,A,B,n);//两两合并
			else
				for (int k = 0; k < n; k++)
					B[k] = A[k];
		}
		printf("%d",B[0]);
		for (int j = 1; j < n; j++)
			printf(" %d",B[j]);
		printf("\n");
	}
	return 0;
}