R2D2 and Droid Army(多棵线段树)

R2D2 and Droid Army

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

An army of n droids is lined up in one row. Each droid is described by m integers a1, a2, …, am, where ai is the number of details of the i-th type in this droid’s mechanism. R2-D2 wants to destroy the sequence of consecutive droids of maximum length. He has m weapons, the i-th weapon can affect all the droids in the army by destroying one detail of the i-th type (if the droid doesn’t have details of this type, nothing happens to it).

A droid is considered to be destroyed when all of its details are destroyed. R2-D2 can make at most k shots. How many shots from the weapon of what type should R2-D2 make to destroy the sequence of consecutive droids of maximum length?

Input

The first line contains three integers n, m, k (1 ≤ n ≤ 105, 1 ≤ m ≤ 5, 0 ≤ k ≤ 109) — the number of droids, the number of detail types and the number of available shots, respectively.

Next n lines follow describing the droids. Each line contains m integers a1, a2, …, am (0 ≤ ai ≤ 108), where ai is the number of details of the i-th type for the respective robot.

Output

Print m space-separated integers, where the i-th number is the number of shots from the weapon of the i-th type that the robot should make to destroy the subsequence of consecutive droids of the maximum length.

If there are multiple optimal solutions, print any of them.

It is not necessary to make exactly k shots, the number of shots can be less.

Sample test(s)

Input

5 2 4

4 0

1 2

2 1

0 2

1 3

Output

2 2

Input

3 2 4

1 2

1 3

2 2

Output

1 3

Note

In the first test the second, third and fourth droids will be destroyed.

In the second test the first and second droids will be destroyed.

**题意:有n*m的矩阵，然后你有k发子弹。现在你可以朝着任意列发射子弹，每一发子弹都会使该列上的数值-1，最小减少到0。

现在问你连续最长的行数，在k发子弹内，使得这些行上的数值全部为0.**

思路:以每一列为一颗线段树(m<6),在查找每一个列的某个区间的最大值和与k进行比较,暴力选出最大的区间,并记录这个区间的各列的区间最大值

#include <set>
#include <map>
#include <list>
#include <stack>
#include <cmath>
#include <queue>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define PI cos(-1.0)
#define RR freopen("input.txt","r",stdin)

using namespace std;

typedef long long LL;

const int MAX = 1e5;

int Tree[6][4*MAX];

int n,m,k;

int QMax[6];

int Ans[6];

void Update(int pos,int L,int R,int site,int s,int ans)//建树
{
    if(L==R)
    {
        Tree[pos][site]=ans;
        return ;
    }
    int mid = (L+R)>>1;
    if(s<=mid)
    {
        Update(pos,L,mid,site<<1,s,ans);
    }
    else
    {
        Update(pos,mid+1,R,site<<1|1,s,ans);
    }
    Tree[pos][site]=max(Tree[pos][site<<1],Tree[pos][site<<1|1]);
}

void Query(int L,int R,int l,int r,int site)//查询
{
    if(l<=L&&r>=R)
    {
        for(int i=1;i<=m;i++)
        {
            QMax[i]=max(QMax[i],Tree[i][site]);
        }
        return ;
    }
    int mid=(L+R)>>1;
    if(l<=mid)
    {
        Query(L,mid,l,r,site<<1);
    }
    if(r>mid)
    {
        Query(mid+1,R,l,r,site<<1|1);
    }
}

int main()
{
    int data;

    scanf("%d %d %d",&n,&m,&k);

    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            scanf("%d",&data);
            Update(j,1,n,1,i,data);
        }
    }
    int l=1,r=1,Max=0;
    bool flag;
    for(r=1;r<=n;r++)
    {
        flag=false;
        while(!flag)
        {
            memset(QMax,0,sizeof(QMax));
            Query(1,n,l,r,1);
            LL sum=0;
            for(int i=1;i<=m;i++)
            {
                sum+=QMax[i];
            }
            if(sum<=k)
            {
                flag=true;
            }
            else
            {
                l++;
            }
        }
        if(Max<(r-l+1))
        {
            memcpy(Ans,QMax,sizeof(Ans));
            Max=(r-l+1);
        }
    }
    for(int i=1;i<=m;i++)
    {
        printf("%d ",Ans[i]);
    }
    return 0;
}