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

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=;

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=;i<n;i++) pq.push(Item(A[i]+B[],));

    for(i=;i<n;i++)

    {

        item=pq.top();pq.pop();

        C[i]=item.s;b=item.b;

        if(b+<n) pq.push(Item(C[i]-B[b]+B[b+],b+));

    }

}

int main()

{

    int n,i,j;

    while(~scanf("%d",&n))

    {

        for(i=;i<n;i++)

        {

            for(j=;j<n;j++) scanf("%d",&A[i][j]);

            sort(A[i],A[i]+n);

        }

        for(i=;i<n;i++)

            merge(A[],A[i],A[],n);

        for(i=;i<n;i++) printf(i?" %d":"%d",A[][i]);

        printf("\n");

    }

    return ;

}

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