uva 11997 priority_queue 应用举例（超省时间！！！）

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3148

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

Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!

题目大意:有K个元素，在每个数组里取出一个元素加起来，可以得到k^k个和。求这些和中最小的k个数

解题思路：大白书第189页，字数太多，我就不敲了。这种写法的复杂度是K*K*logK,暴力遍历的话会是k^K，一定会超时的

/**
   暴力来写时间复杂度为k^k,是难以接受的，利用本题的解题思路复杂度将为K*K*logK
**/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
struct Item
{
    int s,b;//s=A[a]+B[b];这里的a并不重要，因此不保存
    Item(int _s,int _b)
    {
        s=_s;
        b=_b;
    }
    bool operator<(const Item&rhs)const
    {
        return s>rhs.s;
    }
};
void merge(int *A,int *B,int *C,int n)
{
    priority_queue <Item> q;
    for(int i=0;i<n;i++)
       q.push(Item(A[i]+B[0],0));
    for(int i=0;i<n;i++)
    {
        Item item=q.top();q.pop();//取出A[a]+B[b];
        C[i]=item.s;
        int b=item.b;
        if(b+1<n)q.push(Item(item.s-B[b]+B[b+1],b+1));
        //加入A[a]+B[b+1]=s-B[b]+B[b+1];
    }
}
const int maxn=768;
int A[maxn][maxn];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        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 i=1;i<n;i++)
            merge(A[0],A[i],A[0],n);
        printf("%d",A[0][0]);
        for(int i=1;i<n;i++)
            printf(" %d",A[0][i]);
        printf("\n");
    }
    return 0;
}