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

Sample Output

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

#include <bits/stdc++.h>

using namespace std;

#define INF 0x3f3f3f3f

#define LC(x) (x<<1)

#define RC(x) ((x<<1)+1)

#define MID(x,y) ((x+y)>>1)

#define CLR(arr,val) memset(arr,val,sizeof(arr))

#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);

typedef pair<int, int> pii;

typedef long long LL;

const double PI = acos(-1.0);

const int N = 3e5 + 7;

const int M = 2e6 + 1e5 + 7;

struct edge

{

    int to, nxt;

    LL w;

    edge() {}

    edge(int _to, int _nxt, LL _w): to(_to), nxt(_nxt), w(_w) {}

};

edge E[M];

int head[N], tot;

int vis[N];

LL ds[N], de[N], Mindist[N];

void init()

{

    CLR(head, -1);

    tot = 0;

}

inline void add(int s, int t, LL d)

{

    E[tot] = edge(t, head[s], d);

    head[s] = tot++;

}

void spfa(int s, int flag, LL d[])

{

    CLR(vis, 0);

    if (flag)

        CLR(ds, INF), ds[s] = 0;

    else

        CLR(de, INF), de[s] = 0;

    queue<int>Q;

    Q.push(s);

    while (!Q.empty())

    {

        int u = Q.front();

        Q.pop();

        vis[u] = 0;

        for (int i = head[u]; ~i; i = E[i].nxt)

        {

            int v = E[i].to;

            if (d[v] > d[u] + E[i].w)

            {

                d[v] = d[u] + E[i].w;

                if (!vis[v])

                {

                    vis[v] = 1;

                    Q.push(v);

                }

            }

        }

    }

}

int main(void)

{

    int tcase, n, m, si, u, i;

    scanf("%d", &tcase);

    for (int q = 1; q <= tcase; ++q)

    {

        init();

        scanf("%d%d", &n, &m);

        LL ti;

        for (i = 1; i <= m; ++i)

        {

            scanf("%I64d%d", &ti, &si);

            add(i, i + m, ti); //m

            while (si--)

            {

                scanf("%d", &u);

                add(u + (m << 1), i, 0LL); //si

                add(i + m, u + (m << 1), 0LL); //si

            }

        }

        spfa(1 + (m << 1), 1, ds);

        spfa(n + (m << 1), 0, de);

        LL ans = 0x3f3f3f3f3f3f3f3f;

        printf("Case #%d: ", q);

        vector<int>pos;

        for (i = 2 * m + 1; i <= 2 * m + n; ++i)

        {

            Mindist[i] = max<LL>(ds[i], de[i]);

            if (Mindist[i] < ans)

                ans = Mindist[i];

        }

        if (ans == 0x3f3f3f3f3f3f3f3f)

            puts("Evil John");

        else

        {

            printf("%I64d\n", ans);

            for (i = 2 * m + 1; i <= 2 * m + n; ++i)

                if (Mindist[i] == ans)

                    pos.push_back(i - (m << 1));

            int sz = pos.size();

            for (i = 0; i < sz; ++i)

                printf("%d%c", pos[i], " \n"[i == sz - 1]);

        }

    }

    return 0;

}