POJ2442——Sequence

Sequence

Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 8965 Accepted: 2990

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

Source

POJ Monthly,Guang Lin

题意：

给你m(0<m<=100)个序列包含n(0<n<=2000)个正整数。现在每个序列选择一个数并组成一个新的序列，很明显有n^m种序列。现在我们需要计算出这n^m种序列每种序列的数列之和，输出前n个最小的数列之和。

思路：

优先队列，a数组存和，b存该组数据，

运用优先队列，我们可以维持第一个序列到第二个序列前n个最小的数列之和，之后递推到第n个序列。

a[0]+b[0]<=a[0]+b[1]......<=a[0]+b[n]

a[1]+b[0]<=a[1]+b[1]......<=a[1]+b[n]

......

a[2]+b[0]<=a[2]+b[1]......<=a[2]+b[n]

#include <iostream>#include <vector>#include <queue>#include <cstdlib>#include <cstdio>#include <algorithm>using namespace std;int main(){    int T;    cin>>T;    while(T--)    {        int n, m;        cin>>n>>m;        priority_queue<int,vector<int>,less<int> > q;        int *a = new int[m];        int *b = new int[m];        for( int i = 0;i < m;i++ )            cin>>a[i];        sort(a,a+m);        for( int i = 1;i < n;i++ )        {            for( int j = 0;j < m;j++ )            {                cin>>b[j];                q.push(a[0]+b[j]);            }            sort(b,b+m);            for( int j = 1;j < m;j++ )            {                int flag = 0;                //标识变量                for( int k = 0;k < m;k++ )                {                    if( a[j]+b[k]<q.top() )     //当和小于顶部元素才进行更新                    {                        q.pop();                        q.push(a[j]+b[k]);                        flag = 1;                    }                    else                        break;                }                if( !flag )                    break;            }            for( int j = 0;j < m;j++ )    //更新存和数组            {                a[m-j-1] = q.top();                q.pop();            }        }        for( int j = 0;j < m;j++ )        {            if(!j)                cout<<a[j];            else                cout<<" "<<a[j];        }        cout<<endl;    }    return 0;}