POJ-2442 Sequence K路归并问题

　　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。

  1 //STATUS:G++_AC_532MS_768KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 typedef long long LL;
 33 typedef unsigned long long ULL;
 34 //const
 35 const int N=2010;
 36 const int INF=0x3f3f3f3f;
 37 const int MOD=100000,STA=8000010;
 38 const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 int a[N],b[N],temp[N];
 58 int T,m,n;
 59 
 60 struct Node{
 61     int num,a,b;
 62     friend bool operator < (const Node &a,const Node &b){
 63         return a.num>b.num;
 64     }
 65 };
 66 
 67 priority_queue<Node> q;
 68 
 69 int main()
 70 {
 71  //   freopen("in.txt","r",stdin);
 72     int i,j;
 73     Node t;
 74     scanf("%d",&T);
 75     while(T--)
 76     {
 77         scanf("%d%d",&m,&n);
 78         for(i=0;i<n;i++)
 79             scanf("%d",&a[i]);
 80         sort(a,a+n);
 81         for(i=1;i<m;i++){
 82             while(!q.empty())q.pop();
 83             for(j=0;j<n;j++)
 84                 scanf("%d",&b[j]);
 85             sort(b,b+n);
 86             for(j=0;j<n;j++)
 87                 q.push(Node{a[j]+b[0],j,0});
 88             for(j=0;j<n;j++){
 89                 t=q.top();q.pop();
 90                 temp[j]=t.num;
 91                 q.push(Node{a[t.a]+b[t.b+1],t.a,t.b+1});
 92             }
 93             for(j=0;j<n;j++)a[j]=temp[j];
 94         }
 95 
 96         printf("%d",a[0]);
 97         for(j=1;j<n;j++)
 98             printf(" %d",a[j]);
 99         putchar('
');
100     }
101     return 0;
102 }