hdu 5521 最短路

Meeting

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1656    Accepted Submission(s): 515

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.

/*
hdu 5521 最短路

problem:
给你n个点的图,甲在1,乙在n. 它们同时移动,问相聚时的最小花费. 然后是m个点集,点集内的任意两点之间的移动花费为ti

solve:
因为是同时移动. 所以分别对1和n求一个最短路. 然后每个节点取两个最短路中的最大值就能得到花费.
最开始想的是建立所有边,但是边的数量会太多.
所有走到一个节点时,将其所在的所有点集处理一遍. 而且只需要处理一次即可,已经维护了一个最短状态.

hhh-2016-08-30 19:56:48
*/
#pragma comment(linker,"/STACK:124000000,124000000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <vector>
#include <math.h>
#include <queue>
#include <set>
#include <map>
#define lson  i<<1
#define rson  i<<1|1
#define ll long long
#define clr(a,b) memset(a,b,sizeof(a))
#define scanfi(a) scanf("%d",&a)
#define scanfs(a) scanf("%s",a)
#define scanfl(a) scanf("%I64d",&a)
#define key_val ch[ch[root][1]][0]
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
const ll mod = 1e9+7;
const int maxn = 100010;

struct qnode
{
    int v,c;
    qnode(int _v = 0 ,int _c =0) : v(_v),c(_c){}
    bool operator <(const qnode &a)const
    {
        return c > a.c;
    }
};

bool vis[maxn];
bool tvis[maxn];
ll dis1[maxn], dis2[maxn],cost[maxn];
int anspos[maxn];
int num[maxn];
vector<int> pos[maxn];
vector<int> have[maxn];
int T,n,m;
void dijkstra(int start,ll dis[])
{
    for(int i= 1;i <= n;i++)
    {
        vis[i] = tvis[i] = 0;
        dis[i] = inf;
    }
    priority_queue<qnode> q;
    q.push(qnode(start,0));
    dis[start] = 0;
    qnode t;
    while(!q.empty())
    {
        t = q.top();
        q.pop();
        int u = t.v;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = 0 ;i < pos[u].size();i++)
        {
            int t = pos[u][i];
//            if(tvis[t]) continue;
//            tvis[t] = 1;
            for(int j = 0;j<have[t].size();j++)
            {
                int v = have[t][j];
                if(v == u)
                    continue;
                if(dis[v] > dis[u] + cost[t])
                {
                    dis[v] = dis[u] + cost[t];
                    q.push(qnode(v,dis[v]));
                }
            }
        }
    }
}

int main()
{
    int x;
//    freopen("in.txt","r",stdin);
    scanfi(T);
    int cas = 1;
    while(T--)
    {
        scanfi(n),scanfi(m);
        for(int i = 0 ;i <= n;i++)
           pos[i].clear();
        for(int i = 0;i <= m;i++)
            have[i].clear();
        for(int i = 1;i <= m;i++)
        {
            scanf("%I64d%d",&cost[i],&num[i]);
            for(int j = 0;j < num[i];j++)
            {
                scanfi(x);
                pos[x].push_back(i);
                have[i].push_back(x);
            }
        }
        dijkstra(1,dis1);
        dijkstra(n,dis2);
        int cnt = 0;
        ll ans = inf;
        for(int i = 1;i <= n;i++)
        {
            ll t = max(dis1[i],dis2[i]);
            if(t != inf){
                ans = min(ans,t);
            }
        }
        for(int i =1; i <= n;i++)
        {
            if(max(dis1[i],dis2[i]) == ans)
            anspos[cnt++] = i;
        }
        printf("Case #%d: ",cas++);
       if(ans != inf)
       {
           printf("%I64d
",ans);
           for(int i = 0;i < cnt;i++)
           {
               printf("%d%c",anspos[i],i == cnt-1 ? '
':' ');
           }
       }
       else
        printf("Evil John
");
    }
    return 0;
}