Codeforces Round #291 (Div. 2) D. R2D2 and Droid Army RMQ问题 ST算法

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 thei-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.

题意，给一排机器人，每个机器人有m个值，m个炮弹，每个炮弹发射一次，每个相应的值就减1,直到减为0结束，要求总发射次数不超过k次的情况下

连续为0的机器人个数最多，二分枚举长度，对每个长度计算i i+len之间的最大值，相加不超过k,就可以使这len排的数全部减成0,所以问题转化成了rm

q问题 ，用线段树 st算法等等，都可以总复杂度，n*log(n);

#define N 100005#define MOD 1000000000000000007int n,m,k;int maxsum[N][21][6];void RMQ(int num){    for(int mm = 0;mm < m;mm++){        for(int j = 1; j < 20; ++j)            for(int i = 1; i <= num; ++i)                if(i + (1 << j) - 1 <= num)                {                    maxsum[i][j][mm] = max(maxsum[i][j - 1][mm] , maxsum[i + (1 << (j - 1))][j - 1][mm] );                }    }}int getRMQ(int i,int j,int m){    int k=log2( j - i + 1);    return max(maxsum[i][k][m], maxsum[j-(1<<k)+1][k][m]);}int check(int len){    for(int i=1;i+len-1<=n;i++){        int ans = 0;        FJ(m){           ans +=getRMQ(i,i+len-1,j);        }        if(ans <= k)        return i;    }    return -1;}void outPut(int start,int len){    FJ(m){        if(j)            printf(" %d",len==0?0:getRMQ(start,start+len-1,j));        else            printf("%d",len==0?0:getRMQ(start,start+len-1,j));    }    printf("\n");}int main(){    S2(n,m);    {        S(k);        FI(n){            FJ(m){                S(maxsum[i+1][0][j]);            }        }        RMQ(n);        int s = 1,e = n,mid;        while(s<e-1){            mid = (s+e)/2;            if(check(mid) != -1)                s = mid;            else                e = mid;        }        int t = 0;        if((t = check(e))!= -1){            outPut(t,e);        }        else if((t = check(s))!= -1){            outPut(t,s);        }        else {            outPut(t,0);        }    }    return 0;}