Codeforces Round #291 (Div. 2) D. R2D2 and Droid Army [线段树+线性扫一遍]

传送门

D. 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, ..., a_m, where a_i 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, ..., a_m (0 ≤ a_i ≤ 10⁸), where a_i 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.

题解如标题，线段树+线性扫一遍即可~~~

9855946	2015-02-15 10:16:13	njczy2010	D - R2D2 and Droid Army	GNU C++	Accepted	186 ms	27412 KB
9855894	2015-02-15 10:10:30	njczy2010	D - R2D2 and Droid Army	GNU C++	Wrong answer on test 6	15 ms	27400 KB

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<string>

#define N 100005
#define M 1505
//#define mod 10000007
//#define p 10000007
#define mod2 1000000000
#define ll long long
#define LL long long
#define eps 1e-6
//#define inf 2147483647
#define maxi(a,b) (a)>(b)? (a) : (b)
#define mini(a,b) (a)<(b)? (a) : (b)

using namespace std;

ll n,m,k;
ll ma;
ll ans[10];

typedef struct
{
    ll t[7];
}PP;
PP a[N];
PP tree[4*N];

PP build(ll i,ll l,ll r)
{
    //printf(" i=%I64d l=%I64d r=%I64d
",i,l,r);
    if(l==r){
        tree[i]=a[l];
        return tree[i];
    }
    PP le,ri;
    ll mid=(l+r)/2;
    le=build(2*i,l,mid);
    ri=build(2*i+1,mid+1,r);
    ll j;
    for(j=1;j<=m;j++){
        tree[i].t[j]=max(le.t[j],ri.t[j]);
    }
    return tree[i];
}

PP query(ll i,ll l,ll r,ll L,ll R)
{
    //printf(" i=%I64d l=%I64d r=%I64d L=%I64d R=%I64d
",i,l,r,L,R);
    if(l>=L && r<=R) return tree[i];
    ll mid;
    mid=(l+r)/2;
    PP le,ri,re;
    ll j;
    for(j=1;j<=m;j++){
        le.t[j]=ri.t[j]=0;
    }
    if(mid>=L){
        le=query(i*2,l,mid,L,R);
    }
    if(mid<R){
        ri=query(i*2+1,mid+1,r,L,R);
    }
    for(j=1;j<=m;j++){
        re.t[j]=max(le.t[j],ri.t[j]);
    }
    return re;
}

void ini()
{
    memset(ans,0,sizeof(ans));
    ma=0;
    ll i,j;
    for(i=1;i<=n;i++){
        for(j=1;j<=m;j++){
            scanf("%I64d",&a[i].t[j]);
        }
    }
    //printf(" bb
");
    build(1,1,n);
}

void solve()
{
    ll st,en;
    st=1;
    en=0;
    ll now=0;
    PP re;
    ll j;
    for(j=1;j<=m;j++){
        re.t[j]=0;
    }
    while(en<n)
    {
        en++;
        now=0;
        for(j=1;j<=m;j++){
            re.t[j]=max(re.t[j],a[en].t[j]);
            now+=re.t[j];
        }
        while(now>k){
            st++;
            if(st>en){
                for(j=1;j<=m;j++){
                    re.t[j]=0;
                }
                break;
            }
            re=query(1,1,n,st,en);
            now=0;
            for(j=1;j<=m;j++){
                now+=re.t[j];
            }
            //printf(" st=%I64d en=%I64d now=%I64d
",st,en,now);
        }
        if(now>k) continue;
       // printf(" st=%I64d en=%I64d now=%I64d ma=%I64d
",st,en,now,ma);
        if(en-st+1>ma){
            ma=en-st+1;
            for(j=1;j<=m;j++){
                ans[j]=re.t[j];
            }
        }
    }
}

void out()
{
    printf("%I64d",ans[1]);
    ll i;
    for(i=2;i<=m;i++){
        printf(" %I64d",ans[i]);
    }
    printf("
");
}

int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    //scanf("%d",&T);
    //for(int ccnt=1;ccnt<=T;ccnt++)
    //while(T--)
    //scanf("%d%d",&n,&m);
    while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
    {
        ini();
        solve();
        out();
    }
    return 0;
}