Meeting

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 5353    Accepted Submission(s): 1700

Problem Description

Bessie and her friend Elsie decide to have a meeting. However, after Farmer John decorated his fences they were separated into different blocks. John's farm are divided into n blocks labelled from 1 to n. Bessie lives in the first block while Elsie lives in the n-th one. They have a map of the farm which shows that it takes they ti minutes to travel from a block in Ei to another block in Ei where Ei (1≤i≤m) is a set of blocks. They want to know how soon they can meet each other and which block should be chosen to have the meeting.

Input

The first line contains an integer T (1≤T≤6), the number of test cases. Then T test cases follow. The first line of input contains n and m. 2≤n≤105. The following m lines describe the sets Ei (1≤i≤m). Each line will contain two integers ti(1≤ti≤109)and Si (Si>0) firstly. Then Si integer follows which are the labels of blocks in Ei. It is guaranteed that ∑mi=1Si≤106.

Output

For each test case, if they cannot have the meeting, then output "Evil John" (without quotes) in one line. Otherwise, output two lines. The first line contains an integer, the time it takes for they to meet. The second line contains the numbers of blocks where they meet. If there are multiple optional blocks, output all of them in ascending order.

Sample Input

2 5 4 1 3 1 2 3 2 2 3 4 10 2 1 5 3 3 3 4 5 3 1 1 2 1 2

Sample Output

Case #1: 3 3 4 Case #2: Evil John

Hint

In the first case, it will take Bessie 1 minute travelling to the 3rd block, and it will take Elsie 3 minutes travelling to the 3rd block. It will take Bessie 3 minutes travelling to the 4th block, and it will take Elsie 3 minutes travelling to the 4th block. In the second case, it is impossible for them to meet.

思路：不可能兩兩點相互之間建立邊，這樣邊數太多肯定爆，但考慮這些點是在一個集合裡的，這樣對於每一個集合建立一個集合點，所有集合內的點指向這個點，建立兩個有向邊，一個為0,另一個為ti,這樣就可以實現集合內點的連線了，然後跑一個最短路。

程式碼：

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e6+10;

struct ed{
    int  to;
    int next;
    int val;
}e[maxn];
int head[maxn];
int tot;
int n,m;
long long d1[maxn],d2[maxn];
long long d[maxn];
int vis[maxn];
void add(int f,int to,int val){
    e[tot].to=to;
    e[tot].val=val;
    e[tot].next=head[f];
    head[f]=tot++;
}

void SPFA(int s,int nn){
    for(int i=0;i<=nn;i++) d[i]=0x3f3f3f3f,vis[i]=0;
    queue<int >q;
    while(!q.empty()) q.pop();
    q.push(s);
    vis[s]=1;
    d[s]=0;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].to;
            int w=e[i].val;
            if(d[v]>d[u]+w){
                d[v]=d[u]+w;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int main(){
    int t;
    scanf("%d",&t);
    int cas=1;
    while(t--){
            tot=0;
            memset(head,-1,sizeof(head));
            scanf("%d%d",&n,&m);
            for(int i=1;i<=m;i++){
                int t1,t2;
                scanf("%d%d",&t1,&t2);
                for(int j=1;j<=t2;j++){
                    int tem;
                    scanf("%d",&tem);
                    add(tem,n+i,t1);
                    add(n+i,tem,0);
                }
            }

            SPFA(1,n+m+1);
            for(int i=1;i<=n;i++) d1[i]=d[i];
            SPFA(n,n+m+1);
            for(int i=1;i<=n;i++) d2[i]=d[i];

            int mi=0x3f3f3f3f;
            int f=0;
            for(int i=1;i<=n;i++){
                    long long tem=max(d1[i],d2[i]);
//                    printf("--%lld\n",tem);
                    if(mi>tem) f=i,mi=tem;
            }
            printf("Case #%d: ",cas++);
            if(mi==0x3f3f3f3f) printf("Evil John\n");
            else {
                printf("%lld\n",mi);
                printf("%d",f);
                for(int i=1;i<=n;i++) if(max(d1[i],d2[i])==mi&&i!=f) printf(" %d",i);
                printf("\n");
            }
    }
}