POJ-2442 Sequence K路归并问题

　　题目链接：http://poj.org/problem?id=2442

　　问题一：K个有序表合成一个有序表，元素共有n个。用堆优化

　　问题二：两个序列的前n小的元素。堆优化。

　　这题就是问题二的扩展，每次处理两个序列，求出两个序列的前n小的元素，然后把前n小的元素看做一个序列，再和下一个序列一起处理，依次类推下去。

 //STATUS:G++_AC_532MS_768KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int a[N],b[N],temp[N];
 int T,m,n;

 struct Node{
     int num,a,b;
     friend bool operator < (const Node &a,const Node &b){\
         return a.num>b.num;
     }
 };

 priority_queue<Node> q;

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j;
     Node t;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&m,&n);
         for(i=;i<n;i++)
             scanf("%d",&a[i]);
         sort(a,a+n);
         for(i=;i<m;i++){
             while(!q.empty())q.pop();
             for(j=;j<n;j++)
                 scanf("%d",&b[j]);
             sort(b,b+n);
             for(j=;j<n;j++)
                 q.push(Node{a[j]+b[],j,});
             for(j=;j<n;j++){
                 t=q.top();q.pop();
                 temp[j]=t.num;
                 q.push(Node{a[t.a]+b[t.b+],t.a,t.b+});
             }
             for(j=;j<n;j++)a[j]=temp[j];
         }

         printf("%d",a[]);
         for(j=;j<n;j++)
             printf(" %d",a[j]);
         putchar('\n');
     }
     return ;
 }