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
#include
#include
#include
#include
using namespace std;
int m,n;
int t[55000][7];
int time[2][55000];
int main()
{
scanf("%d%d",&m,&n);
for(int i=1;i<=m;i++)
{
for(int j=1;j<=n;j++)
{
scanf("%d",&t[i][j]);
}
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
time[i%2][j]=max(time[i%2][j-1]+t[j][i],time[(i-1)%2][j]+t[j][i]);
}
}
for(int i=1;i<=m;i++)
printf("%d ",time[n%2][i]);
return 0;
}
C. Booking System
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Innovation technologies are on a victorious march around the planet. They integrate into all spheres of human activity!
A restaurant called "Dijkstra's Place" has started thinking about optimizing the booking system.
There are n booking requests received by now. Each request is characterized by two numbers: ci and pi —
the size of the group of visitors who will come via this request and the total sum of money they will spend in the restaurant, correspondingly.
We know that for each request, all ci people
want to sit at the same table and are going to spend the whole evening in the restaurant, from the opening moment at 18:00 to the closing moment.
Unfortunately, there only are k tables in the restaurant. For each table, we know ri —
the maximum number of people who can sit at it. A table can have only people from the same group sitting at it. If you cannot find a large enough table for the whole group, then all visitors leave and naturally, pay nothing.
Your task is: given the tables and the requests, decide which requests to accept and which requests to decline so that the money paid by the happy and full visitors was maximum.
Input
The first line of the input contains integer n (1?≤?n?≤?1000)
— the number of requests from visitors. Then n lines follow. Each line contains two integers: ci,?pi (1?≤?ci,?pi?≤?1000)
— the size of the group of visitors who will come by the i-th request and the total sum of money they will pay when they visit the restaurant, correspondingly.
The next line contains integer k (1?≤?k?≤?1000) —
the number of tables in the restaurant. The last line contains k space-separated integers: r1,?r2,?...,?rk (1?≤?ri?≤?1000) —
the maximum number of people that can sit at each table.
Output
In the first line print two integers: m,?s — the number of accepted requests and the total money you get from these requests, correspondingly.
Then print m lines — each line must contain two space-separated integers: the number of the accepted request and the number of the table to seat people who
come via this request. The requests and the tables are consecutively numbered starting from 1 in the order in which they are given in the input.
If there are multiple optimal answers, print any of them.
Sample test(s)
input
3
10 50
2 100
5 30
3
4 6 9
output
2 130
2 1
3 2
#include
#include
#include
#include
using namespace std;
int n,m;
struct VS
{
int c,p,id,tb;
}vit[1100];
bool cmp(VS a,VS b)
{
if(a.p!=b.p) return a.p>b.p;
return a.cmxpeople) continue;
for(int j=0;jr[j].cp) continue;
if(r[j].tk==0)
{
r[j].tk=1;
vit[i].tb=r[j].id;
ans++; sum+=vit[i].p;
break;
}
}
}
cout<
