POJ 2442-Sequence（優先佇列-m組n個數每組取一個求n個最小值）

kewlgrl發表於2017-01-31
佇列

Sequence
Time Limit: 6000MS   Memory Limit: 65536K
Total Submissions: 9285   Accepted: 3097

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

Source

POJ Monthly,Guang Lin

題目意思：

有m組序列，每組有n個數。從每組中取一個數並求和，輸出最小的n個和。


解題思路：

輸入一組處理一組，先求出前兩組數的n個最小和，再輸入一組數，將這組數與前面的n個最小和一起處理，再求出這兩組數的n個最小和…………直到m組數全部輸入處理完畢。

處理過程如下：

①第一組數輸入A陣列中後降序排列，第二組數輸入B陣列後也降序排列

②將A[0]+B[i]這n個數先入隊，再依次遍歷A陣列剩下的n-1個數與B陣列的n個數之和，每次比較將較小的和入隊。

③遍歷完畢後，將佇列中元素賦給A陣列，同時佇列清空。

④輸入下一組數到B陣列，重複②，直到m組數輸入處理完畢。


#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
const int MAXN=30030;
int a[MAXN],b[MAXN];
using namespace std;
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        priority_queue<int>q;//最大值優先
        int m,n;
        cin>>m>>n;
        for(int i=0; i<n; ++i)
            cin>>a[i];
        sort(a,a+n);//升序排列
        --m;
        while(m--)
        {
            for(int i=0; i<n; ++i)
                cin>>b[i];
            sort(b,b+n);
            for(int i=0; i<n; ++i)//入隊
                q.push(a[0]+b[i]);
            for(int i=1; i<n; ++i)//依次比較將較小的和入隊
                for(int j=0; j<n; ++j)
                {
                    int temp=a[i]+b[j];
                    if(temp<q.top())
                    {
                        q.pop();
                        q.push(temp);
                    }
                }
            for(int i=0; i<n; ++i)//將佇列中的數賦給a陣列，同時佇列清空
            {
                a[i]=q.top();
                q.pop();
            }
            sort(a,a+n);
        }
        for(int i=0; i<n-1; ++i)//輸出結果
            cout<<a[i]<<" ";
        cout<<a[n-1]<<endl;
    }
    return 0;
}
/*
1
2 3
1 2 3
2 2 3
*/

測試資料：

2
2 3
1 2 3
2 2 3
10 10
21 12 123 3 21 123 32 143 43 56
2 32 43 34 54 56 656 76 43 234
234 45 5 65 56 76 43 23 435 57
32 324 435 46 56 76 87 78 43 23
3 32 324 45 56 57 34 23 54 565
23 32 34 342 324 232 2 432 324 12
234 324 4 45 65 67 435 23 5 654
34 3245 345 56 56 657 67 456 345 325
234 234 546 65 88 66 53 654 65 765
5 3 34 34 34 56 345 234 2 34

執行結果：

3 3 4
131 132 132 133 134 135 140 140 141 141


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