Bus Pass_BFS***

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 970    Accepted Submission(s): 400

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

思路：假设任意一个地区做为中心区，然后计算其到各个车站区的距离，保存其最大值。算完所有区域后，最大值最小的区域就是中心区。 

逆向思维：以车站区为出发点，计算其到地图上全部可到达区域的距离值。这样一下子就可以减轻很大计算量（起始位置从全部区域减少为所有车站区）

 用两个队列来实现（记为q[0]和q[1]），因为算法需要像水波一样逐层向外层推进，需要先把紧邻当前的一圈地区都处理完之后，在处理更外层的一个圈。这种水波推进的实现方法，其实就是交替对两个队列q[0],q[1]进行操作，即在处理q[0]中元素时，把下一步要入列的都放进q[1]中，这样就达到了隔离的效果。

#include<iostream>
#include<string.h>
#include<queue>
#include<stdio.h>
using namespace std;
const int inf=0x7777777;
const int N=10005;
int num[N],reach[N],edge[N][10],res[N],cur;//edge[i][j]表示标号为i的地区的第j个相邻区域标号；
//reach[i]==cur表示地区i在cur+1站中已被访问；num[i]记录标号为i的地区的相邻地区个数；res[i]记录每条线路上每个地区带地区i距离中的最大值
void bfs(int s)
{
    queue<int>qu[2];
    int a=0,b=1,val=0;
    if(reach[s]<cur)
    {
        qu[b].push(s);
        reach[s]=cur;
        res[s]=max(res[s],val+1);
    }
    while(!qu[b].empty())
    {
        swap(a,b);//通过a，b在两个队列之间灵活切换；
        val++;
        while(!qu[a].empty())
        {
            int t=qu[a].front();
            qu[a].pop();
            for(int i=0;i<num[t];i++)
            {
                if(reach[edge[t][i]]<cur)
                {
                    qu[b].push(edge[t][i]);
                    reach[edge[t][i]]=cur;
                    res[edge[t][i]]=max(res[edge[t][i]],val+1);
                }
            }
        }
    }
}

int main()
{
    int t;
    int n,m,id,mm;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        memset(reach,-1,sizeof(reach));
        memset(res,0,sizeof(res));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&id);
            scanf("%d",&num[id]);
            for(int j=0;j<num[id];j++)
            {
                scanf("%d",&edge[id][j]);
            }
        }
        cur=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d",&mm);
            for(int j=0;j<mm;j++)
            {
                scanf("%d",&id);
                bfs(id);
                cur++;
            }
        }
        int ret=inf,center=-1;
        for(int i=0;i<10000;i++)
        {
            if(reach[i]==cur-1&&res[i]<ret)
            {
                center=i;
                ret=res[i];
            }
        }
        printf("%d %d
",ret,center);
    }
    return 0;
}