Sequence
Time Limit: 6000MS | Memory Limit: 65536K | |
Total Submissions: 8887 | Accepted: 2953 |
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
题目大意:给你m个序列,每次从每个序列里挑选一个数组成一个新的序列,每个序列进行加和组成一个新的序列,然后输出前n小的序列。
对于两个序列我们就可以a[0]+b[0],a[1]+b[0]一直到a[n-1]+b[0],然后是依次加a[1],这样我们就可以得到一个新的序列,如果只有两个序列,那我们的序列就是符合条件的,如果对于三个序列那就是两个序列组成的新序列看成新的a序列,第三个序列就是b序列,想一想,是不是?
这样依次迭代,到最后就是所求的m个序列组成的新序列,如果我们全部暴力存储,显然时间和空间开销都是非常大的,到最后跟直接m个序列暴力是没有区别的,达到n^m。
╮(╯▽╰)╭人家也没让你求辣么多啊,每次维护长为n的数组就可以啦
这里我用一个优先队列来维护哒464ms才
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <queue>
using namespace std;
const int MAXN=2000+10;
const int inf=0x3f3f3f;
int a[MAXN],b[MAXN],c[MAXN];
int n,m;
void del()
{
int i,j;
priority_queue<int>q;
for(i=0;i<n;++i)//先读入n个元素
{
q.push(a[i]+b[0]);
}
for(i=1;i<n;++i)//维护队列
{
for(j=0;j<n;++j)
{
int x=a[j]+b[i];
if(x<q.top())//如果比最大值还小,就进队,大的出队
{
q.pop();
q.push(x);
}
else break;//否则的话,后面一定都比x大的,所以直接break
}
}
for(i=0;i<n;++i)
{
a[n-1-i]=q.top();//更新a数组
q.pop();
}
}
int main()
{
int t;
int i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&m,&n);
for(j=0;j<n;++j)scanf("%d",&a[j]);
sort(a,a+n);
for(i=1;i<m;++i)
{
for(j=0;j<n;++j)scanf("%d",&b[j]);
sort(b,b+n);
del();
}
for(i=0;i<n-1;++i)printf("%d ",a[i]);
printf("%d\n",a[n-1]);
}
return 0;
}