STL--F - Sequence（n*m->之前的最低要求m个月）

F - Sequence
Time Limit:6000MS Memory Limit:65536KB 64bit IO Format:%I64d
& %I64u
Submit Status

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

1
2 3
1 2 3
2 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 <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
int a[110][2100] , b[2100] ;
priority_queue <int> 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;
}