poj 2442 Sequence

题目链接：点击打开链接

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

题目大意：有M行　每行有n个数，每行拿出一个数来相加得和，共有n的m次方个和，求前N个最小的和

基本思路：用优先队列存储　每输入一行就对其进行加和判断存储，注意删数队列；

具体看代码

<span style="font-size:24px;">#include <iostream>#include<queue>#include<cstring>#include<algorithm>#include<cstdlib>#include<cstdio>using namespace std;int main(){    int t;    int n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&m,&n);        int x;        priority_queue <int ,vector<int >,greater<int> >q;//从小到达存数        priority_queue <int ,vector<int >,less<int > >p;//从大到小存储        int a[2005];        for(int i = 0; i<n; i++)        {            scanf("%d",&x);//先将一行存入队列            q.push(x);        }        for(int i=1; i<m; i++)        {            for(int j=0; j<n; j++)            {                scanf("%d",&a[j]);            }            while(!q.empty())//将上一行的每个数拿出与当前行加和            {                int sum=q.top();                q.pop();                for(int j=0;j<n;j++)                {                    if(p.size()>=n&&p.top()>sum+a[j])//因为只要前n个数　所以只保留前n个最小的就行                    {                        p.pop();                        p.push(sum+a[j]);                    }                    else if(p.size()<n)                    {                        p.push(sum+a[j]);                    }                }            }            while(!p.empty())            {                q.push(p.top());                p.pop();            }        }        for(int i=0;i<n;i++)        {            if(i==0)            {                printf("%d",q.top());                q.pop();            }            else            {                printf(" %d",q.top());                q.pop();           }        }        printf("\n");    }    return 0;}</span>