HDU 5521.Meeting 最短路模板题

Meeting

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3361    Accepted Submission(s): 1073

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.

 

Source

2015ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5521

题意：有m个集合，每个集合里面的任意两点均有一条距离为ei的无向边，求1和n到其他点的最短距离中最大值的最小值。

思路：最短路模板题。每个集合作为作为一个点，对应的点到集合的距离为ei，最后答案/2。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<bitset>
#include<queue>
#include<stack>
#include<map>
#include<vector>
using namespace std;
#define eps 0.0000001
typedef long long ll;
typedef pair<int,int> P;
const int maxn=2e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;
const ll INF=1e18+7;
struct edge
{
    int from,to;
    ll w;
};
vector<edge>G[maxn];
priority_queue<P,vector<P>,greater<P> >q;
ll dist[2][maxn];
void addedge(int u,int v,ll w)
{
    G[u].push_back((edge)
    {
        u,v,w
    });
    G[v].push_back((edge)
    {
        v,u,w
    });
}
void dij(int t,int s)
{
    dist[t][s]=0LL;
    q.push(P(dist[t][s],s));
    while(!q.empty())
    {
        P p=q.top();
        q.pop();
        int u=p.second;
        for(int i=0; i<G[u].size(); i++)
        {
            edge e=G[u][i];
            if(dist[t][e.to]>dist[t][u]+e.w)
            {
                dist[t][e.to]=dist[t][u]+e.w;
                q.push(P(dist[t][e.to],e.to));
            }
        }
    }
}
void init(int n)
{
    for(int i=0; i<=2*n+10; i++) G[i].clear();
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int Case=1; Case<=T; Case++)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=m; i++)
        {
            int val;
            scanf("%lld",&val);
            int t;
            scanf("%d",&t);
            while(t--)
            {
                int s;
                scanf("%d",&s);
                addedge(s,n+i,val);
            }
        }
        for(int i=0; i<=2*n+10; i++) dist[0][i]=dist[1][i]=INF;
        dij(0,1);
        dij(1,n);
        ll ans=INF;
        for(int i=1; i<=n; i++)
        {
            //printf("%lld %lld
",dist[0][i],dist[1][i]);
            ans=min(ans,max(dist[0][i],dist[1][i]));
        }
        printf("Case #%d: ",Case);
        if(ans>=INF) puts("Evil John");
        else
        {
            printf("%lld
",ans/2);
            int cou=0;
            for(int i=1; i<=n; i++)
            {
                if(!cou&&max(dist[0][i],dist[1][i])==ans) printf("%d",i),cou++;
                else if(cou&&max(dist[0][i],dist[1][i])==ans) printf(" %d",i),cou++;
            }
            printf("
");
        }
        init(n);
    }
    return 0;
}