uva 11997 优先队列

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

题目大意：给k个数组各含k个整数，求每个数组中取一个元素他们的和最小的前k个。
分析：一共有k的k次方个和，先看只有两个数组的情况(从小到大排好序的)
     1       2      3    ...
1  A1+B1<=A1+B2<=A1+B3.....
2  A2+B1<=A2+B2<=A2+B3.....
...........................
若计算两两之和需k的平方次不可取（时间复杂度O（n的平方））
用优先队列找出最小的前k个和（时间复杂度O（nlogn））
由上面的表格知，第一行的k个元素在它所在列中是和最小的，但不是最小的前k个
把第一行的元素放入优先队列中，把队列的顶部元素提取出来，它是队列中最小的(A[a]+B[b])
也是两两之和中未被提取出来的最小的，把与它最相近的即比它大或等于它的（A[a]+B[b+1]）
元素放入队列中。
以此类推提取出来的k个元素是最小的前k个。
两两数组合并从而求得最后结果。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;

const int Max=755;
int A[Max][Max];

struct Item
{
    int s,b;
    Item() {}
    Item(int s,int b):s(s),b(b) {}
    bool operator < (const Item &a)const{
        return a.s<s;
    }
}item;

void merge(int *A,int *B,int *C,int n)
{
    priority_queue<Item> pq;
    int i,b;
    for(i=0;i<n;i++) pq.push(Item(A[i]+B[0],0));
    for(i=0;i<n;i++)
    {
        item=pq.top();pq.pop();
        C[i]=item.s;b=item.b;
        if(b+1<n) pq.push(Item(C[i]-B[b]+B[b+1],b+1));
    }
}
int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++) scanf("%d",&A[i][j]);
            sort(A[i],A[i]+n);
        }
        for(i=1;i<n;i++)
            merge(A[0],A[i],A[0],n);
        for(i=0;i<n;i++) printf(i?" %d":"%d",A[0][i]);
        printf("
");
    }
    return 0;
}