STL--F - Sequence（n*m->之前的最低要求m个月）-白红宇

F - Sequence

Time Limit:6000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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Description

Given m sequences, each contains n non-negative integer. Now we may select one number from each sequence to form a sequence with m integers. It's clear that we may get n ^ m this kind of sequences. Then we can calculate the sum of numbers in each sequence, and get n ^ m values. What we need is the smallest n sums. Could you help us?

Input

The first line is an integer T, which shows the number of test cases, and then T test cases follow. The first line of each case contains two integers m, n (0 < m <= 100, 0 < n <= 2000). The following m lines indicate the m sequence respectively. No integer in the sequence is greater than 10000.

Output

For each test case, print a line with the smallest n sums in increasing order, which is separated by a space.

Sample Input

12 31 2 32 2 3

Sample Output

3 3 4

感谢http://www.cnblogs.com/372465774y/archive/2012/07/09/2583866.html

做这个题首先思考两个问题

由这两个得出，要求n个数组每一个数组m个值。数组1和数组2的和找出最小的m个，再用来和数组3求和，找到最小的m个，终于得到全部的数组中的最小的m个

因为每一个数组都是有序的，并且我们要求的最小的m个。数组a[i][j]+队列中的值 > 队首的值，那么a[i][j]加上队列中以后的值都会大于队首。对于我们要求解的最小的m个值无意义。队列中保存了当前数组到之前全部数组的最小的m个和，不断更新队列

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;int a[110][2100] , b[2100] ;priority_queue 
        
          p ;int main(){    int i , j , k , n , m , t ;    scanf("%d", &t);    while(t--)    {        scanf("%d %d", &n, &m);        for(i = 0 ; i < n ; i++)        {            for(j = 0 ; j < m ; j++)                scanf("%d", &a[i][j]);            sort(a[i],a[i]+m);        }        for(i = 0 ; i < m ; i++)            p.push(a[0][i]) ;        for(i = 1 ; i < n ; i++)        {            for(j = 0 ; j < m ; j++)            {                b[j] = p.top();                p.pop();            }            for(j = 0 ; j < m ; j++)            {                for(k = m-1 ; k >= 0 ; k--)                {                    if(j == 0)                        p.push( a[i][j]+b[k] );                    else                    {                        if( a[i][j] + b[k] < p.top() )                        {                            p.pop();                            p.push(a[i][j]+b[k]);                        }                        else                            break;                    }                }            }        }        for(j = 0 ; j < m ; j++)        {            b[j] = p.top();            p.pop();        }        for(j = m-1 ; j >= 0 ; j--)        {            if(j == 0)                printf("%d\n", b[j]);            else                printf("%d ", b[j]);        }    }    return 0;}

本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/4731914.html，如需转载请自行联系原作者