Exam Problem schedule matrix

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yousef Yousef
yousef Yousef 2014 年 4 月 7 日
コメント済み: yousef Yousef 2014 年 4 月 7 日
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 件のコメント
yousef Yousef
yousef Yousef 2014 年 4 月 7 日
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 2014 年 4 月 7 日

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