Exam Problem schedule matrix

4 vues (au cours des 30 derniers jours)
yousef Yousef
yousef Yousef le 7 Avr 2014
Commenté : yousef Yousef le 7 Avr 2014
Let’s say, the first column represents the trip number in a train station. The other columns represent the number of passengers coming from different gates (g1 g2 g3...g10). So if the trip number is not repeated (which means there is only one trip a day), then put all passengers in one train. If the number is repeated (two shifts or more, in the morning, other in the evening & last at night).Take the passengers from 3 gates which have the largest number of passengers in the morning, other 3 in the evening & last group at night.
end
Trip # g1 g2 g3 g4 g5 g6 g7 g8 g9 g10
2 82 10 16 15 66 76 71 83 44 49
4 91 28 98 43 4 75 4 70 39 45
4 13 55 96 92 85 40 28 32 77 65
1 92 96 49 80 94 66 5 96 80 71
1 64 97 81 96 68 18 10 4 19 76
The result should look like this:
2:
83 82 76 71 66 49 44 16 15 10
8 1 6 7 5 10 9 3 4 2
4, 1:
98 91 75 70 45
3 1 6 8 10
4, 2:
43 39 28 4 4
4 9 2 5 7
1, 4:
96 96 94 92 80
2 8 5 1 4
1, 5:
80 71 66 49 5
9 10 6 3 7
  2 commentaires
yousef Yousef
yousef Yousef le 7 Avr 2014
I want to know what is wrong with this solution
finalresult=[];
for i=1:4
if true
% code
end
for j=i+1:5
result1=[];
result2=[];
result3=[];
if w(j)~=w(i)
[a,b]=sort(r(i,:),'descend');
presult1=b;
else
[c,d]=sort(r(j,:),'descend');
presult2=d(1:3);
presult3=d(4:6);
end
result1=[result1;presult1];
result2=[result2;presult2];
result3=[result3;presult3];
end
end
if true
% code
end
yousef Yousef
yousef Yousef le 7 Avr 2014

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