POJ-2442 Sequence(手写堆优化)

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

#include "bits/stdc++.h"
#define mem(a,b) memset(a,b,sizeof(a))
#define F(z,x,y) for (z=x;z<=y;z++)
using namespace std;
typedef long long LL;
const int MAX=2005;
int cas;
int n,m;
int s[MAX],sum[MAX];
struct Que{
    int h[MAX*100];
    int n;
    Que (){
        mem(h,0);
        n=0;
    }
    void heapify(int x){
        int child=x*2,key=h[x];
        while (child<=n){
            if (child<n && h[child]<h[child+1]) child++;
            if (key<h[child]){h[x]=h[child];x=child,child=x*2;}
            else break;
        }
        h[x]=key;
    }
    void insert(int key){
        int x=++n;
        while (x>1){
            if (key>h[x/2]) h[x]=h[x/2],x/=2;
            else break;
        }
        h[x]=key;
    }
    void del(){
        if (n==1) n=0;
        else h[1]=h[n--],heapify(1);
    }
};
int main(){
    freopen ("sequence.in","r",stdin);
    freopen ("sequence.out","w",stdout);
    int i,j,k;
    scanf("%d",&cas);
    while (cas--){
        scanf("%d%d",&n,&m);
        mem(s,0),mem(sum,0);
        Que q;
        F(i,1,m){
            scanf("%d",s+i);
        }
        F(i,2,n){
            F(j,1,m)
             scanf("%d",sum+j);
            F(j,1,m)
             q.insert(sum[1]+s[j]);
            F(j,2,m)
             F(k,1,m){
                 int mx=sum[j]+s[k];
                 if (mx<q.h[1]){
                     q.del();
                     q.insert(mx);
                 }
             }
            F(j,1,m)
             s[j]=q.h[j];
            mem(q.h,0);
            q.n=0;
        }
        sort(s+1,s+m+1);
        F(i,1,m){
            printf("%d ",s[i]);
        }
        printf("
");
    }
    return 0;
}