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

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, ..., 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 44 01 22 10 21 3
Output
2 2
Input
3 2 41 21 32 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.

 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