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 }