codeforces 416B

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 tijis 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.

Sample test(s)

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 

题目大意：

有m副画，有n个画家，每个画家画完第一幅画之后才可以画第二幅画，每一副画需要n个画家来完成，只有第一个画家完成了，第二个画家才可以开始（随便谁开始画都行，第一个画家全部画完画或者画几张然后第2个画家再开始画也可以，），求每幅画完成时的最短时间是多少

思路

这是一道dp的题目，dp[i][j]表示第i副画第j个画家完成时的最短时间

状态转移方程：

dp[i][j]=max(dp[i][j-1],dp[i-1][j])+a[i][j];

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define M 100005
int a[M][15];
int dp[M][15];
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&a[i][j]);//i代表画，j代表画家
            }
        }
        memset(dp,0,sizeof(dp));
        int tmp=0;
        for(int i=1;i<=n;i++)
        {
            tmp+=a[1][i];
            dp[1][i]=tmp;
        }
        tmp=0;
        for(int i=1;i<=m;i++)
        {
            tmp+=a[i][1];
            dp[i][1]=tmp;
        }
        for(int i=2;i<=m;i++)
        {
            for(int j=2;j<=n;j++)
            {
                dp[i][j]=max(dp[i][j-1],dp[i-1][j])+a[i][j];
            }
        }
        for(int i=1;i<m;i++)
        {
            printf("%d ",dp[i][n]);
        }
        printf("%d\n",dp[m][n]);
    }
    return 0;
}