UVA 11997 K Smallest Sums

 

Problem K

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

 

题意：每行选一个加在一起，则总共有k^k个和，求前k小的值都是多少。

解析：如果排序的话必然超时。所以算法就是先把每行都升序排好，两行两行的求前k小的值，求出来后赋值入数组再继续求下去。即分成a0+b0,a0+b1,a0+b2,,,;a1+b0,a1+b1,,,;a2+b0,a2+b1,a2+b2;则最小的值一定在a0+b0,a1+b0,a2+b0,,,之间，例如最小的为a1+b0，那么就把它踢出队列，在加入队列a1+b1,继续上述操作，直到找出前k小。

#include
#include
#include
#include
#define INF 800
using namespace std;

struct numt
{
	friend bool operator < (numt n1,numt n2)//优先队列比较结构的固定模版
	{
		return n1.sum > n2.sum;//此处>代表从小到大排列，若为<则是从大到小
	}
	int sum,n;
};
int k,num[INF][INF];
int ans[INF];

void fx(int *a,int *b)	//a,b两个数列求前k最小值
{
	priority_queueQ;
	int cnt = 0;
	numt temp;
	for(int i = 0; i < k ; i ++)	//先赋初始值
	{
		temp.sum = a[i]+b[0];
		temp.n = 0;
		Q.push(temp);
	}
	for(int i = 1; i < k ; i ++)
	{
		numt t1 = Q.top();
		Q.pop();
		a[i-1] = t1.sum ;		//最小的都赋给a,因为a在后面都没用到了，直接赋值
		t1.sum = t1.sum+b[t1.n+1]-b[t1.n];
		t1.n ++;
		Q.push(t1);
	}
	a[k-1] = Q.top().sum;
}

int main()
{
	while(scanf("%d",&k) != EOF)
	{
		for(int i = 0 ; i < k ; i ++)
		{
			for(int j = 0 ;j < k ; j++)
			{
				scanf("%d",&num[i][j]);
			}
			sort(num[i],num[i]+k);	//一定要先排序好
		}
		for(int i = 1; i < k ; i ++)
		{
			fx(num[0],num[i]);
		}
		printf("%d",num[0][0]);
		for(int i = 1 ; i < k; i ++)
			printf(" %d",num[0][i]);
		printf("\n");
	}
	return 0;
}