HDU 2377 Bus Pass

Bus Pass

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1007    Accepted Submission(s): 430

Problem Description

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 cetera.

You have a list of all bus trips you frequently make, and would 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.

nz lines starting with two integers id_i (1 <= id_i <= 9 999) and mz_i (1 <= mz_i <= 10), a number identifying the i-th zone and the number of zones adjacent to it, followed by mz_i integers: the numbers of the adjacent zones.

nr lines starting with one integer mr_i (1 <= mr_i <= 20), indicating the number of zones the ith bus trip visits, followed by mr_i integers: 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 7406

7401 6 7412 7402 7400 7406 7410 7411

7402 5 7412 7403 7400 7401 7411

7403 6 7413 7414 7404 7400 7402 7412

7404 5 7403 7414 7415 7405 7400

7405 6 7404 7415 7407 7408 7406 7400

7406 7 7400 7405 7407 7408 7409 7410 7401

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 7410

7412 6 7416 7411 7401 7402 7403 7413

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

 

题意：一个城市有很多区，提供的多条公交线，求每个区到公交线上的区的最长距离中最短的，但是如果有这个距离相等的取区号最小的。

 

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int inf=100000;
int n,m;
int dis[210][10010];
bool flag[10010];
int point[10010],num[10010],d[10010][210];
void spfa(int st,int cou){
    queue<int> q;
    memset(flag,false,sizeof(flag));
    while(!q.empty()){
        q.pop();
    }
    flag[st]=true;
    q.push(st);
    for(int i=1;i<=n;i++){
        dis[cou][point[i]]=inf;
    }
    dis[cou][st]=0;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=1;i<=num[x];i++){
            int y=d[x][i];
            if(!flag[y]){
                q.push(y);
                dis[cou][y]=dis[cou][x]+1;
                flag[y]=true;
            }
        }
    }
}
int main(){
    cin.sync_with_stdio(false);
    int T;
    cin>>T;
    while(T--){
        cin>>n>>m;
        for(int i=1;i<=n;i++){
            cin>>point[i];
            cin>>num[point[i]];
            for(int j=1;j<=num[point[i]];j++){
                cin>>d[point[i]][j];
            }
        }
        int nn,a;
        int k=1;
        while(m--){
            cin>>nn;
            while(nn--){
                cin>>a;
                spfa(a,k);
                k++;
            }
        }
        int value=10000,kk;
        for(int i=1;i<=n;i++){
            int now=0;
            for(int j=1;j<=k;j++){
                if(dis[j][point[i]]>now){
                    now=dis[j][point[i]];
                }
            }
            if(value>now){
                value=now;
                kk=point[i];
            }
            else if(value==now&&kk>point[i]){
                kk=point[i];
            }
        }
        cout<<(value+1)<<" "<<kk<<endl;
    }
    return 0;
}

2017-01-24 13:35:55