Sequence
| Time Limit: 6000MS | Memory Limit: 65536K | |
| Total Submissions: 9316 | Accepted: 3113 |
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
题意
给出m组数，每组n个，求从每一组挑选出一个数相加产生前n小的和。
思路
假设只有一组数，那么前n小的和便是这组的所有元素，如果增加一组数之后，只有原来组的前n小的和才会对新组产生贡献（因为其他数相加产生的和一定不会是结果所说的前n个里面的），于是用一个二重循环来枚举计算添加完这一组之后的前n小的和。
AC 代码
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>
typedef __int64 LL;
int num[2001];
priority_queue<int,vector<int>,greater<int> >d; //小顶堆
priority_queue<int>q; //大顶堆
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int m,n,x;
scanf("%d%d",&m,&n);
for(int i=0; i<n; i++)
{
scanf("%d",&x);
d.push(x); //首先当只有一组的时候
}
for(int i=1; i<m; i++) //添加剩下的组
{
for(int j=0; j<n; j++)
scanf("%d",num+j);
while(!d.empty())
{
x=d.top();
d.pop();
for(int j=0; j<n; j++) //二重循环计算
{
if((int)q.size()<n)
q.push(x+num[j]);
else if(x+num[j]<q.top())
{
q.pop();
q.push(x+num[j]);
}
}
}
while(!q.empty())
{
d.push(q.top());
q.pop();
}
}
printf("%d",d.top());
d.pop();
while(!d.empty())
{
printf(" %d",d.top());
d.pop();
}
printf("\n");
}
return 0;
}
