POJ2442 Sequence(堆的骚操作)

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

题意:有m行数字,每行均有n个数字.从每行均取出一个数字相加,共有m^n个解,输出解最小的n个.

题解:建一个大根堆,先将第一行的全部扔进去,在全部扔回一个数组里,那么此时数组中的数是从大到小排列的.

然后这个数组,我们给它中的每一个数取出,再加上第二行的第一个数,全部扔进堆里,接着对于第2~n个数中的每一个,我们将它与数组中最小的,次小的……(数组倒过来)相加,如果和比大根堆的堆顶大,这就是我们要找的了,将堆顶弹出,将这个值插入.然而我并不知道这玩意的复杂度,居然就这么A了?真是奇怪,有信仰的话可一定要用手写堆啊!

代码如下:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define hi puts("hi");
using namespace std;

int map[105][2005];

struct Heap
{
    int a[300010],sz;

    void init()
    {
        sz=1;
    }

    void add(int x)
    {
        int now=sz++;
        while(now>1)
        {
            int fa=now>>1;
            if(a[fa]>x)
            {
                break;
            }
            a[now]=a[fa];
            now=fa;
        }
        a[now]=x;
    }

    int pop()
    {
        int ans=a[1];
        int now=1;
        int x=a[--sz];
        while((now<<1)<sz)
        {
            int ls=now<<1,rs=now<<1|1;
            if(a[ls]<a[rs]&&rs<sz)
            {
                ls=rs;
            }
            if(a[ls]<x)
            {
                break;
            }
            a[now]=a[ls];
            now=ls;
        }
        a[now]=x;
        return ans;
    }

    int top()
    {
        return a[1];
    }

    int empty()
    {
        return sz==1;
    }
} heap;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int m,n,b[200010],cnt=0;
        heap.init();
        scanf("%d%d",&m,&n);

        for(int i=1; i<=m; i++)
        {
            for(int j=1; j<=n; j++)
            {
                scanf("%d",&map[i][j]);
            }
        }
        for(int i=1; i<=n; i++)
        {
            heap.add(map[1][i]);
        }
        for(int i=2; i<=m; i++)
        {
            cnt=0;
            while(!heap.empty())
            {
                b[++cnt]=heap.pop();
            }
            for(int j=1; j<=cnt; j++)
            {
                heap.add(b[j]+map[i][1]);
            }
            for(int j=2; j<=n; j++)
            {
                for(int k=cnt; k>=1; k--)
                {
                    if(b[k]+map[i][j]<heap.top())
                    {
                        heap.pop();
                        heap.add(b[k]+map[i][j]);
                    }
                    else
                    {
                        break;
                    }
                }
            }
        }
        cnt=0;
        while(!heap.empty())
        {
            b[++cnt]=heap.pop();
        }
        for(int i=cnt; i>=2; i--)
        {
            printf("%d ",b[i]);
        }
        printf("%d
",b[1]);
    }

}