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 nn blocks labelled from 11 to nn. 
Bessie lives in the first block while Elsie lives in the nn-th one. They have a map of the farm 
which shows that it takes they titi minutes to travel from a block in EiEito another block 
in EiEi where Ei (1≤i≤m)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.

InputThe first line contains an integer T (1≤T≤6)T (1≤T≤6), the number of test cases. Then TT test cases 
follow.

The first line of input contains nn and mm. 2≤n≤1052≤n≤105. The following mm lines describe the sets Ei (1≤i≤m)Ei (1≤i≤m). Each line will contain two integers ti(1≤ti≤109)ti(1≤ti≤109) and Si (Si>0)Si (Si>0) firstly. Then SiSi integer follows which are the labels of blocks in EiEi. It is guaranteed that ∑mi=1Si≤106∑i=1mSi≤106.OutputFor 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

可以将给顶集合的元素连到一个虚拟结点上，求出最短路来再/2,这样避免了大量的重复加边，还避免了小数

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

#define INF 0x3f3f3f3f

/*
连接虚拟结点
到该点的距离为L
求出最短路/2 避免小数！
*/
LL T, d, n, m, cnt;
struct edge
{
edge(LL _a,LL _b):to(_a),cost(_b){}
LL to, cost;
};
vectorE[MAXN];
LL dist1[MAXN], dist2[MAXN];
bool vis[MAXN];
void addedge(LL f,LL to,LL dis)
{
E[f].push_back(edge(to, dis));
E[to].push_back(edge(f, dis));
}
void init()
{
for (LL i = ; i < MAXN; i++) E[i].clear(); } void spfa(LL beg, LL lowcost[]) { queue q;
memset(vis, false, sizeof(vis));
for (int i = ; i <= n + m; i++) lowcost[i] = INF; lowcost[beg] = ; vis[beg] = true; q.push(beg); while (!q.empty()) { LL f = q.front(); q.pop(); vis[f] = false; for (int i = ; i < E[f].size(); i++) { if (lowcost[E[f][i].to] > lowcost[f] + E[f][i].cost)
{
lowcost[E[f][i].to] = lowcost[f] + E[f][i].cost;
if (!vis[E[f][i].to])
{
vis[E[f][i].to] = true;
q.push(E[f][i].to);
}
}
}
}
}
int main()
{
scanf("%lld", &T);
for(LL cas = ;cas <= T; cas++)
{
init();
scanf("%lld%lld", &n, &m);
LL tmp, tt;
for (LL i = ; i <= m; i++)
{
scanf("%lld%lld", &d, &tmp);
while (tmp--)
{
scanf("%lld", &tt);
addedge(tt, n + i, d);
}
}
spfa(, dist1);
spfa(n , dist2);
LL ans = INF;
for (int i = ; i <= n; i++)
ans = min(ans, max(dist1[i], dist2[i]));
if (ans == INF)
printf("Case #%lld: Evil John\n", cas);
else
{
printf("Case #%lld: %lld\n", cas, ans / );
bool f = false;
for (int i = ; i <= n; i++)
{
if (max(dist1[i], dist2[i]) == ans)
{
if (!f)
printf("%d", i), f = true;
else
printf(" %d", i);
}
}
printf("\n");
}
}
}