Zoj 2913 bus pass (BFS)

Bus Pass Time Limit: 5 seconds memory limit: 32768 KB

You travel a lot by bus and the costs of all the Seperate tickets are starting to add up.

Therefore you want to see if it might be advantageous for you to buy a bus pass.

The way the bus system works in your country (and also in the Netherlands) is as follows:

When you buy a bus pass, you have to indicate a center zone and a star value. you are allowed to travel freely in any zone which has a distance to your center zone which is less than your star value. for example, if you have a star value of one, you can only travel in your center zone. if you have a star value of two, you can also travel in all adjacent zones, et Cees.

You have a list of all bus trips you frequently make, and wowould like to determine the minimum star value you need to make all these trips using your buss pass. but this is not always an easy task. for example look at the following figure:

 

Here you want to be able to travel from A to B and from B to D. the best center zone is 7400, for which you only need a star value of 4. note that you do not even visit this zone on your trips!

 

Input

On the first line an integerT(1 <=T<= 100): the number of test cases. Then for each test case:

One line with two integersNZ(2 <=NZ<= 9 999) andNR(1 <=NR<= 10): The number of zones and the number of bus trips, respectively.

NZLines starting with two integersIDI(1 <= Idi <= 9 999) andMZI(1 <=MZI<= 10), a number identifying the I-th zone and the number of zones adjacent to it, followed by MZI integers: the numbers of the adjacent zones.

NRLines starting with one integerMRI(1 <=MRI<= 20), indicating the number of zones the ith bus trip visits, followedMRIIntegers: the numbers of the zones through which the bus passes in the order in which they are visited.

All zones are connected, either directly or via other zones.

Output

For each test case:

One line with two integers, the minimum star value and the ID of a center zone which achieves this minimum star value. If there are multiple possibilities, choose the zone with the lowest number.

Sample Input

1
17 2
7400 6 7401 7402 7403 7404 7405
7401 6 7412 7402 7400 7406 7410
7402 5 7412 7403 7400 7401
7403 6 7413 7414 7404 7400 7402
7404 5 7403 7414 7415 7405
7405 6 7404 7415 7407 7408 7406
7406 7 7400 7405 7407 7408 7409 7410
7407 4 7408 7406 7405 7415
7408 4 7409 7406 7405 7407
7409 3 7410 7406 7408
7410 4 7411 7401 7406 7409
7411 5 7416 7412 7402 7401
7412 6 7416 7411 7401 7402 7403
7413 3 7412 7403 7414
7414 3 7413 7403 7404
7415 3 7404 7405 7407
7416 2 7411 7412
5 7409 7408 7407 7405 7415
6 7415 7404 7414 7413 7412 7416

Sample output

4 7400

 

 

I have been doing this question for a few days. I asked someone else to teach me a thought. It is actually the opposite of the correct answer. The result can be imagined. However, I have also listened well and deepened my understanding of this question. Later I went to the Internet to read other people's ideas and code for solving problems. I copied it.

Note: If the memset function is used to assign a value to the array, It is theoretically acceptable. However, if the value of threshold1 is submitted here, it is wrong. The value must be assigned using. Be careful with the memset function in the future.

It means you need to find the threshold between the points and each bus station as small as possible, and output the maximum threshold value of the node bus station.

Solution: Find the minimum distance between each bus online point and each point, and use it to compare with the threshold value of each new point (of course, the maximum threshold value ), then traverse the entire threshold and find the minimum threshold.

 

# Include <stdio. h> # include <string. h >#include <queue> # define maxn 10000 # define INF 0x3f3f3fusing namespace STD; int edge [maxn] [10], MZ [maxn]; int threshold1 [maxn], threshold2 [maxn]; // threshold1 is used to store the threshold value, and threshold2 is used to store each updated void BFS (int id) {int top; For (INT I = 0; I <= maxn; I ++) threshold2 [I] = inf; queue <int> q; threshold2 [ID] = 1; q. push (ID); While (! Q. empty () {Top = Q. front (); q. pop (); For (INT I = 0; I <MZ [Top]; I ++) {If (threshold2 [edge [Top] [I]> threshold2 [Top] + 1) // find the minimum threshold value from the online point to each point {threshold2 [edge [Top] [I] = threshold2 [Top] + 1; q. push (edge [Top] [I]) ;}}for (INT I = 0; I <= maxn; I ++) // update the maximum threshold value of each vertex {If (threshold1 [I] <threshold2 [I] & threshold2 [I]! = Inf) {threshold1 [I] = threshold2 [I] ;}} int main () {int t; int X, Y; int NZ, NR, Mr, ID; scanf ("% d", & T); While (t --) {for (INT I = 0; I <= maxn; I ++) threshold1 [I] =-1; scanf ("% d", & NZ, & nr); For (INT I = 0; I <NZ; I ++) {scanf ("% d", & ID); scanf ("% d", & MZ [ID]); For (Int J = 0; j <MZ [ID]; j ++) {scanf ("% d", & edge [ID] [J]) ;}}for (INT I = 0; I <NR; I ++) {scanf ("% d", & Mr); For (Int J = 0; j <Mr; j ++ ){ Scanf ("% d", & ID); BFS (ID) ;}} x = inf; For (INT I = 0; I <= maxn; I ++) // find the minimum threshold {If (x> threshold1 [I] & threshold1 [I]! =-1) {x = threshold1 [I]; y = I ;}} printf ("% d \ n", x, y);} return 0 ;}