| # | Problem | Pass Rate (passed user / total user) |
|---|---|---|
| 11248 | Eating Group |
|
| 11250 | The Cost to Pass the Forest |
|
Description
[Partial Judge: C]
There are N hungry people, numbered from 1 to N. Each of them is going to have a lunch with someone else or alone.
With the given input, try to find out that which restaurant to go for each person. Each restaurant is also represented by an integer.
Notice:
You need to implement all three functions defined in the 11248.h header file.
There are two arrays of length 100 (a,b) leaved for you to implement your method. That should be enough.
Read input hint:
int i, a;
scanf("%d" , &a);
for ( i=0; i<N-1; i++) {
scanf(",%d" , &a);
...
}
Input
The input contains three lines.
First line only has a single number N (N<=100), representing the total number of hungry people.
Second line is a comma-seperated sequence of N integers. The ith integer indicates who will ith person eat with. 0 means unknown.
Third line is also a comma-seperated sequence of N integers. The ith integer indicates which restaurant will ith person eat at. 0 means unknown. No '\n' at the end of this line.
Note, ith entry in second line is zero iff ith entry in third line is non-zero.
Output
Output a line of N comma saperated integers with no trailing \n behind.
Each integer indicates the restaurant to eat for the corresponding person.
Sample Input Download
Sample Output Download
Partial Judge Code
11248.cPartial Judge Header
11248.hTags
Discuss
Description
The forest is described as follows:
S: the start of the forest
T: the destination of the forest
#: the passway in the forest
*: the trees in the forest
Now please count the cost to pass through the forest.
Input
The first line of the input give two integer M N, representing the height and the width of the forest.
Then the following M*N arrays gives the description of the forest.
Note. 3<=M, N<=100
Output
Give the minimum cost(steps) to pass through the forest.
If the destination is unreachable, the cost will be 0.
Remember to print a '\n' at the end of the output.