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

传送门

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

Output

2 2

Input

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

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<stack>
 11 #include<string>
 12 
 13 #define N 100005
 14 #define M 1505
 15 //#define mod 10000007
 16 //#define p 10000007
 17 #define mod2 1000000000
 18 #define ll long long
 19 #define LL long long
 20 #define eps 1e-6
 21 //#define inf 2147483647
 22 #define maxi(a,b) (a)>(b)? (a) : (b)
 23 #define mini(a,b) (a)<(b)? (a) : (b)
 24 
 25 using namespace std;
 26 
 27 ll n,m,k;
 28 ll ma;
 29 ll ans[10];
 30 
 31 typedef struct
 32 {
 33     ll t[7];
 34 }PP;
 35 PP a[N];
 36 PP tree[4*N];
 37 
 38 PP build(ll i,ll l,ll r)
 39 {
 40     //printf(" i=%I64d l=%I64d r=%I64d\n",i,l,r);
 41     if(l==r){
 42         tree[i]=a[l];
 43         return tree[i];
 44     }
 45     PP le,ri;
 46     ll mid=(l+r)/2;
 47     le=build(2*i,l,mid);
 48     ri=build(2*i+1,mid+1,r);
 49     ll j;
 50     for(j=1;j<=m;j++){
 51         tree[i].t[j]=max(le.t[j],ri.t[j]);
 52     }
 53     return tree[i];
 54 }
 55 
 56 PP query(ll i,ll l,ll r,ll L,ll R)
 57 {
 58     //printf(" i=%I64d l=%I64d r=%I64d L=%I64d R=%I64d\n",i,l,r,L,R);
 59     if(l>=L && r<=R) return tree[i];
 60     ll mid;
 61     mid=(l+r)/2;
 62     PP le,ri,re;
 63     ll j;
 64     for(j=1;j<=m;j++){
 65         le.t[j]=ri.t[j]=0;
 66     }
 67     if(mid>=L){
 68         le=query(i*2,l,mid,L,R);
 69     }
 70     if(mid<R){
 71         ri=query(i*2+1,mid+1,r,L,R);
 72     }
 73     for(j=1;j<=m;j++){
 74         re.t[j]=max(le.t[j],ri.t[j]);
 75     }
 76     return re;
 77 }
 78 
 79 void ini()
 80 {
 81     memset(ans,0,sizeof(ans));
 82     ma=0;
 83     ll i,j;
 84     for(i=1;i<=n;i++){
 85         for(j=1;j<=m;j++){
 86             scanf("%I64d",&a[i].t[j]);
 87         }
 88     }
 89     //printf(" bb\n");
 90     build(1,1,n);
 91 }
 92 
 93 void solve()
 94 {
 95     ll st,en;
 96     st=1;
 97     en=0;
 98     ll now=0;
 99     PP re;
100     ll j;
101     for(j=1;j<=m;j++){
102         re.t[j]=0;
103     }
104     while(en<n)
105     {
106         en++;
107         now=0;
108         for(j=1;j<=m;j++){
109             re.t[j]=max(re.t[j],a[en].t[j]);
110             now+=re.t[j];
111         }
112         while(now>k){
113             st++;
114             if(st>en){
115                 for(j=1;j<=m;j++){
116                     re.t[j]=0;
117                 }
118                 break;
119             }
120             re=query(1,1,n,st,en);
121             now=0;
122             for(j=1;j<=m;j++){
123                 now+=re.t[j];
124             }
125             //printf(" st=%I64d en=%I64d now=%I64d\n",st,en,now);
126         }
127         if(now>k) continue;
128        // printf(" st=%I64d en=%I64d now=%I64d ma=%I64d\n",st,en,now,ma);
129         if(en-st+1>ma){
130             ma=en-st+1;
131             for(j=1;j<=m;j++){
132                 ans[j]=re.t[j];
133             }
134         }
135     }
136 }
137 
138 void out()
139 {
140     printf("%I64d",ans[1]);
141     ll i;
142     for(i=2;i<=m;i++){
143         printf(" %I64d",ans[i]);
144     }
145     printf("\n");
146 }
147 
148 int main()
149 {
150     //freopen("data.in","r",stdin);
151     //freopen("data.out","w",stdout);
152     //scanf("%d",&T);
153     //for(int ccnt=1;ccnt<=T;ccnt++)
154     //while(T--)
155     //scanf("%d%d",&n,&m);
156     while(scanf("%I64d%I64d%I64d",&n,&m,&k)!=EOF)
157     {
158         ini();
159         solve();
160         out();
161     }
162     return 0;
163 }

posted on 2015-02-15 15:19 njczy2010 阅读(275) 评论(0) 编辑收藏举报