【POJ 2442】 Sequence

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXM 110
#define MAXN 2010

int T,n,m,i,j;
int a[MAXM][MAXN],ans[MAXN];

struct info
{
        int p,q,val;
        bool last;
        friend bool operator < (info a,info b)
        {
                return a.val > b.val;
        }
} ;

inline void _merge(int *a,int *b)
{
        int i,j;
        priority_queue< info > q;
        static int res[MAXN];
        info tmp;
        while (!q.empty()) q.pop();
        q.push((info){1,1,a[1]+b[1],false});
        for (i = 1; i <= n; i++)
        {
                tmp = q.top();
                q.pop();
                res[i] = tmp.val;
                q.push((info){tmp.p,tmp.q+1,a[tmp.p]+b[tmp.q+1],true});
                if (!tmp.last) q.push((info){tmp.p+1,tmp.q,a[tmp.p+1]+b[tmp.q],false});
        }
        for (i = 1; i <= n; i++) a[i] = res[i];
}

int main() 
{
        
        scanf("%d",&T);
        while (T--)
        {
                scanf("%d%d",&m,&n);
                for (i = 1; i <= m; i++)
                {
                        for (j = 1; j <= n; j++) scanf("%d",&a[i][j]);    
                        sort(a[i]+1,a[i]+n+1);
                }
                for (i = 1; i <= n; i++) ans[i] = a[1][i];    
                for (i = 2; i <= m; i++) _merge(ans,a[i]); 
                for (i = 1; i < n; i++) printf("%d ",ans[i]);
                printf("%d\n",ans[n]);
        }
        
        return 0;
    
}