Magina, also known as Anti-Mage, is a very cool hero in DotA (Defense of the Ancient).

If you have no idea of him, here is some brief introduction of his legendary antecedents:
Twin sons to the great Prophet, Terrorblade and Magina were blessed with divine powers: Terrorblade granted with an unnatural affinity with life forces; Magina gifted with energy manipulation. Magina's eventual overexposure to the magic gradually augmented
his elemental resistances and bestowed him the unique ability to move faster than light itself. Now, broken by Terrorblade's fall to the dark side, Magina answers the Sentinel's call in a desperate bid to redeem his brother. Every bitter strike turns the Scourge's
evil essences upon themselves, culminating in a finale that forces his enemy to awaken to the void within and spontaneously implode.
Magina has a very IMBA (imbalanced) skill – Blink, yes, move from one place to another place in a wink. Our problem begins at there.
As a formidable hero in the later stage, Magina always farm with the wild monsters for a long time. To make the farming more efficient, Magina use Blink frequently to jump here and there. Here we assume Blink skill has no CD, that is, we can use this skill
at any time we want.
There are N spots of the wild monsters, and Magina can choose any one to begin. For every spots, Magina may use Ti time to kill the monsters and gain Gi units money, or he choose blink to other spots, which is known to our brilliant Magina. If the monsters
in a spot were killed, it won’t appear any more.
Now Magina want to get M units money to but some excellent equipment, say Battle Fury for example. As a hero to save the world, there is no much time left for Magina, he wonders the minimum time for him to gain at least M units money.

The first line contains a single integer T, indicating the number of test cases.
Each test case begins with two integers N, M. Their meanings are the same as the description.
Then N blocks follow, each one describes a spot of wild monsters.
The first line of each block is there integers Ti, Gi and Ki. Ti is the time, Gi is the units of money, Ki is the number of spots Magina can blink to from here.
Then Ki integer Cij follow, indicating the spots’ ID Magina can blink to. You may assume no ID would appear twice.
The spots are described with ID increasing from 1 to N. Input ensure if you can blink from i to j, you can also blink from j to i.

Technical Specification

1. 1 <= T <= 50
2. 1 <= N <= 50
3. 1 <= Ti <= 10000000
4. 1 <= M, Gi <= 1000000000
5. 1 <= Ki < N
6. 1 <= Cij <= N

For each test case, output the case number first, then the minimum time for Magina to gain at least M units money, if can’t, output “Poor Magina, you can't save the world all the time!”.

3
1 4
2 5 0
1 5
1 4 0
4 10
1 9 0
3 3 1
3
3 3 2
2 4
4 4 1
3

Case 1: 2
Case 2: Poor Magina, you can't save the world all the time!
Case 3: 10

题目大意：题目源自Dota，前面一堆介绍敌法师，最后一段才开始说有n堆野兽，每堆野兽屠杀完完需要花费ti时间，可以增加金钱gi，敌法师有瞬移技能，可以从某堆野兽移到另一堆野兽，题目给定相连的点。最后问求最少得到金钱m的情况下的最少时间。

解题思路：用深搜找出连通子图，然后在连通子图中的节点进行屠杀，每堆野兽有两种选择：杀获不杀.这样问题就转为01背包。
     但本题的金钱和屠杀时间范围都特别大，屠杀时间的范围是1-1000万，金钱的范围是1-10亿，不能用常规的两重循环模拟。其实用两个优先队列就可以模拟01背包的计算过程，第一次将（0，0）入队，然后出队转移到下一个状态，转移完这两个状态都进入队列，可以用两个优先队列进行操作，一个表示上一轮的计算结果，一个是这一轮,就相当于滚动数组。在用优先队列模拟的过程中要注意剪枝，如果不剪或者剪得不好就是MLE。只有一种可以，那就是在每次转移完要把队列1复制到队列0的时候，判断下入队的这个点是不是满足价值比前面入队的点金钱少且耗时少，如果是，这个可能比前面入队的那个优，如果不是，这个一定不须再加入队列，因为当前这个状态钱比进入队列的钱少，且用时相等或多，那么进入一个节点时进行更新一定不必先进队列的状态优。比如，（6，4）表示打到钱为6，且耗时4，如果下一个要进队的是(5,5)，那肯定是前面的更优，对吧？就是这样。
    
#include<stdio.h>
#include<iostream>
#include<vector>
#include<queue>
#include<string.h>
using namespace std;

const int inf=0xfffffff;

struct node
{
    int t,g;
    friend bool operator<(node a,node b)
    {
        if(a.g==b.g)
            return a.t>b.t;
        return a.g<b.g;
    }
};

int vist[55],M,ans;
vector<int>tmap[55];
priority_queue<node>q[2];
node val[55];

void dfs(int x)
{
    node p,tp;
    vist[x]=1;

    while(!q[0].empty())
    {
        p=q[0].top();
           q[0].pop();
        q[1].push(p);

        p.g+=val[x].g;
        p.t+=val[x].t;

        if(p.g>=M)
        {
            if(ans>p.t)ans=p.t; continue;
        }
        if(ans<=p.t)continue;

        q[1].push(p);
    }

    tp.t=ans;
    while(!q[1].empty())
    {
        p=q[1].top();
          q[1].pop();
        if(tp.t>p.t)
            q[0].push(p),tp=p;
    }

    for(int i=0;i<tmap[x].size();i++)
    {
        int j=tmap[x][i];
        if(vist[j])
            continue;
        dfs(j);
    }
}

int main()
{
    int T,N,k,u;
    node p;

    scanf("%d",&T);
    for(int c=1;c<=T;c++)
    {
        scanf("%d%d",&N,&M);
        for(int i=1;i<=N;i++)
        {
            tmap[i].clear();
            vist[i]=0;

            scanf("%d%d%d",&val[i].t,&val[i].g,&k);
            while(k--)
            {
                scanf("%d",&u); tmap[i].push_back(u);
            }
        }

        ans=inf;
        for(int s=1;s<=N;s++)
        if(vist[s]==0)
        {
            while(!q[0].empty()) q[0].pop();
            while(!q[1].empty()) q[1].pop();
            p.g=p.t=0;
            q[0].push(p);

            dfs(s);
        }

        printf("Case %d: ",c);
        if(ans!=inf)
        printf("%d\n",ans);
        else
        printf("Poor Magina, you can't save the world all the time!\n");
    }
}