tput different 亞洲 key rate cstring 重現 not divide

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亞洲區沈陽站-重現賽（感謝東北大學） 題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=5521 題意：有m個集合，每個集合裏面的任意兩點均有一條距離為ei的無向邊，求1和n到其他點的最短距離中最大值的最小值。 思路：最短路模板題。每個集合作為作為一個點，對應的點到集合的距離為ei，最後答案/2。 代碼：

#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\n",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\n",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("\n");
        }
        init(n);
    }
    return 0;
}

最短路模板題

HDU 5521.Meeting 最短路模板題