POJ 2442-sequence (priority queue-M group N number of each group takes one to find n minimum value)

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 of the sequence to form a sequence with M integers. It's clear that we could get n ^ m this kind of sequences. Then we can calculate the sum of numbers in each sequence, and get n ^ m values. What are we need is the smallest n sums. Could help us?

Input The first line is a integer T, which shows the number of test cases, and then T test cases follow. The first line of each case contains integers m, n (0 < M <=, 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 was 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
topic Meaning:

There are m sets of sequences with n number per group. Takes a number from each group and sums it, outputting the smallest n sum.

Problem Solving Ideas:

Input a set of processing a group, first to find the first two sets of the number of n minimum and then enter a set of numbers, the number of the group with the previous n minimum and together, and then find the two sets of the number of n minimum and ... All inputs are processed until the M-group count is complete.

The process is as follows:

① the first set of numbers in the input a array, followed by descending order, the second set of input B array is also descending order

② will a[0]+b[i] This n number first queue, and then go through a array of the remaining n-1 number and B array of n number of the sum, each comparison will be smaller and queued.

After the ③ has been traversed, the elements in the queue are assigned to the A array, and the queue is emptied.

④ Enter the next set of numbers to the B array, repeating the ② until the M-number input processing is complete.

#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;//maximum precedence int m,n;
        cin>>m>>n;
        for (int i=0; i<n; ++i) cin>>a[i];
        Sort (a,a+n);//ascending order--m;
            while (m--) {for (int i=0; i<n; ++i) cin>>b[i];
            Sort (b,b+n);
            for (int i=0; i<n; ++i)//Queue Q.push (A[0]+b[i]);  for (int i=1; i<n; ++i)//compare smaller and queue 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)//assigns the number of queues to a array, while the queue empties {a[i]=q.top ();
            Q.pop ();
        } sort (a,a+n);
        } for (int i=0; i<n-1; ++i)//output result cout<<a[i]<< "";
    cout<<a[n-1]<<endl;
} return 0;
 }/* 1 2 3 1 2 3 2 2 3 * *

test Data:

2
2 3
1 2 3
2 2 3
ten
123 3 123-143
2-656
234 45  5-435-324 435-34-342-3-324-32-324----
23 432 324
234 324 4
234 234 546 65 88 66 53 65 435 (5 654) 3245 345 4 765
5 3 34 34 34 56 345 234 2 34

Operation Result:

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