Codeforces Round #241 (Div. 2) B dp

B. Art Union

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

A well-known art union called "Kalevich is Alive!" manufactures objects d'art (pictures). The union consists of n painters who decided to organize their work as follows.

Each painter uses only the color that was assigned to him. The colors are distinct for all painters. Let's assume that the first painter uses color 1, the second one uses color 2, and so on. Each picture will contain all these n colors. Adding the j-th color to the i-th picture takes the j-th painter tij units of time.

Order is important everywhere, so the painters' work is ordered by the following rules:

Each picture is first painted by the first painter, then by the second one, and so on. That is, after the j-th painter finishes working on the picture, it must go to the (j + 1)-th painter (if j < n);

each painter works on the pictures in some order: first, he paints the first picture, then he paints the second picture and so on;

each painter can simultaneously work on at most one picture. However, the painters don't need any time to have a rest;

as soon as the j-th painter finishes his part of working on the picture, the picture immediately becomes available to the next painter.

Given that the painters start working at time 0, find for each picture the time when it is ready for sale.

Input
The first line of the input contains integers m, n (1 ≤ m ≤ 50000, 1 ≤ n ≤ 5), where m is the number of pictures and n is the number of painters. Then follow the descriptions of the pictures, one per line. Each line contains n integers ti1, ti2, ..., tin (1 ≤ tij ≤ 1000), where tij is the time the j-th painter needs to work on the i-th picture.

Output
Print the sequence of m integers r1, r2, ..., rm, where ri is the moment when the n-th painter stopped working on the i-th picture.

Examples

Input

5 1
1
2
3
4
5

Output

1 3 6 10 15

Input

4 2
2 5
3 1
5 3
10 1

Output

7 8 13 21

题意： n幅 m个画家 每个画家按照次序画 给出每个画家画每幅画的时间
对于同一幅画  画家s画完 画家s+1才能画
画家只能同一时间段画一幅画 当结束当前画后 就可以进行下幅画  或  等待下一幅画被前一个画家画完才能画
问每一幅画被最后一个画家画完的时刻

题解： 动态规划处理 dp[i][j] 表示 第j个画家画完第i幅画的时刻
mp[i][j]代表 第j个画家画完第i幅画所需要的时间
转移方程 ：
dp[i][j]=max(dp[i][j-1],dp[i-1][j])+mp[i][j]

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
#define ll long long
#define mod 1e9+7
#define PI acos(-1.0)
#define bug(x) printf("%%%%%%%%%%%%%",x);
#define inf 1e8
using namespace std;
int n,m;
int mp[50005][6];
int time[50005][6];
int main()
{
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
scanf("%d",&mp[i][j]);
}
memset(time,0,sizeof(time));
time[1][1]=mp[1][1];
for(int j=2;j<=m;j++)
time[1][j]=time[1][j-1]+mp[1][j];
for(int j=2;j<=n;j++)
time[j][1]=time[j-1][1]+mp[j][1];
for(int j=2;j<=m;j++)
for(int i=2;i<=n;i++)
{
for(int j=2;j<=m;j++)
{
if(time[i-1][j]>time[i][j-1])
time[i][j]=time[i-1][j]+mp[i][j];
else
time[i][j]=time[i][j-1]+mp[i][j];
}
}
printf("%d",time[1][m]);
for(int i=2;i<=n;i++)
printf(" %d",time[i][m]);
printf("\n");
return 0;
}