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Here is my code. I already pasted input test. Can anyone point out my mistake?

#include <iostream>

using namespace std;

int N;
int A[25], B[25], C[25], D[25][25];

int main()
{
cin >> N;
int i, j, ans;
for (i = 1; i <= N; i++)
{
int k;
cin >> k;
A[k] = i;
}

for (i = 1; i <= N; i ...

Man, I have the same problem.
I used cin.peek first and get WA. It is AC after changing to cin >> N.

Thank so much. However don't know what is going on.

Check this post http://online-judge.uva.es/board/viewto ... e35982d203

This problem is not hard but anyone know the way to speed up the program. The fastest record is 0.000s. How did he do?

I did not see your code but actually I misunderstand this problem. You are right about repeated cell. Cell can be repeated.

As I think, this is a graph problem and you can use BFS to solve.

I think you refer to the out size of the array.

int Path[100][100];

You can only use Path[0][0] to Path[99][99].

When C = 100. See the following code (for example)

for(i=1;i<=C;i++) Path =0;

You assign Path[100][100] --> Run Time Error.

You should change to: Path[101][101]. It will be okie ...

The problem do not say the cell can be repeated or not. But to find the minimal number of moves, it is wise when each cell will be visited no more once.

I think BFS or DFS will work in this problem. If you stil have WA, please post your code. Maybe someone can point out your mistake.

Nothing is wrong with your code. I submit your code and get AC.

I got your idea. Thanks a lot.

How to get the answer for one-dimension with O(n). I can't find any.

Hi,

According to this problem, you will use the route 1 - 2 - 4 - 7. And the maximum number of passengers is 25. But only 24 tourists can go each time because the tourist guide has to ride the bus with each group . The input sample has 99 tourists. With 4 trips only 96 tourists can go to the ...

Thank guys. I have the same mistake. I got AC now.

1
9
8

Can anyone point out my mistake? Thank.


#include <iostream>
#include <string>

using namespace std;

long Num, M, N, numB, numC;
int A[27], B[27], C[27];

int main()
{
cin >> Num;
for (long k = 1; k <= Num; k++)
{
cin >> M >> N;

long i;

for (i = 1; i <= M; i++)
{
string city1, city2 ...