POJ-2442 Sequence(手写堆优化)

Posted 2020-08-17 可惜没如果=_=

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 9131		Accepted: 3037

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

 

首先这题是可以用系统给的priority_queue的，不过毕竟现在我在学手写堆，于是就给自己找麻烦咯~

 

思路： 因为，要每行都取一个，构成一个和sum。需要找出n个sum。  我们需要一行一行的找，不妨先设前两行的最小的n个sum是由第一行n个数和第二行的第一个数构成的。 存入c[]数组里，然后一次遍历第二行其他的数，看是否有小于c[]数组里最大的数，然后替换，这是前两行的最小的n个sum已经找到，存到了c[]数组， 然后找前三行，同样设让第三行的第一个数与数组c[]所有的数加和， 然后遍历第三行其他的数，看是否有小于c[]数组里最大的数。替换。以此类推。找到第m-1行  结束寻找，这时数组c[]里就是题目要求的数组。

思路来源:http://www.tuicool.com/articles/VjuYFn  

程序压了一点行，可能可读性会差一点

 1 #include "bits/stdc++.h"
 2 #define mem(a,b) memset(a,b,sizeof(a))
 3 #define F(z,x,y) for (z=x;z<=y;z++)
 4 using namespace std;
 5 typedef long long LL;
 6 const int MAX=2005;
 7 int cas;
 8 int n,m;
 9 int s[MAX],sum[MAX];
10 struct Que{
11     int h[MAX*100];
12     int n;
13     Que (){
14         mem(h,0);
15         n=0;
16     }
17     void heapify(int x){
18         int child=x*2,key=h[x];
19         while (child<=n){
20             if (child<n && h[child]<h[child+1]) child++;
21             if (key<h[child]){h[x]=h[child];x=child,child=x*2;}
22             else break;
23         }
24         h[x]=key;
25     }
26     void insert(int key){
27         int x=++n;
28         while (x>1){
29             if (key>h[x/2]) h[x]=h[x/2],x/=2;
30             else break;
31         }
32         h[x]=key;
33     }
34     void del(){
35         if (n==1) n=0;
36         else h[1]=h[n--],heapify(1);
37     }
38 };
39 int main(){
40     freopen ("sequence.in","r",stdin);
41     freopen ("sequence.out","w",stdout);
42     int i,j,k;
43     scanf("%d",&cas);
44     while (cas--){
45         scanf("%d%d",&n,&m);
46         mem(s,0),mem(sum,0);
47         Que q;
48         F(i,1,m){
49             scanf("%d",s+i);
50         }
51         F(i,2,n){
52             F(j,1,m)
53              scanf("%d",sum+j);
54             F(j,1,m)
55              q.insert(sum[1]+s[j]);
56             F(j,2,m)
57              F(k,1,m){
58                  int mx=sum[j]+s[k];
59                  if (mx<q.h[1]){
60                      q.del();
61                      q.insert(mx);
62                  }
63              }
64             F(j,1,m)
65              s[j]=q.h[j];
66             mem(q.h,0);
67             q.n=0;
68         }
69         sort(s+1,s+m+1);
70         F(i,1,m){
71             printf("%d ",s[i]);
72         }
73         printf("\n");
74     }
75     return 0;
76 }