【HDOJ5521】Meeting（最短路）

题意：有n个点，m个点集，每个点集中有e[i]个点，同一点集的点互相之间到达需要t[i]单位的时间，求min(max(dis(1,i),dis(i,n)))，i属于[1,n]

输出最小值并増序输出所有使答案取到最小值的i，无解输出Evil John

n<=1e5,1<=t[i]<=1e9,sigma e[i]<=1e6

思路：出烂了的裂点最短路

对于每个点集另外裂一个点id，设该点集中的点为a[i]

a[i]——>id 长度为t[i]

id——>a[i] 长度为0

以1和n为起点跑两遍dijkstra+堆

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N  1200000
 #define M  2200000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)
 #define oo 1ll<<60
 priority_queue<pair<ll,int> > q;

 ll dis1[N],dis2[N];
 int head[N],vet[M],len[M],nxt[M],vis[N],c[N],id,tot;

 int add(int a,int b,int c)
 {
     //printf("%d %d %d\n",a,b,c);
     nxt[++tot]=head[a];
     vet[tot]=b;
     len[tot]=c;
     head[a]=tot;
 }

 ll min(ll x,ll y)
 {
     if(x<y) return x;
     return y;
 }

 ll max(ll x,ll y)
 {
     if(x>y) return x;
     return y;
 }

 int main()
 {
     freopen("M.in","r",stdin);
     freopen("M.out","w",stdout);
     int cas;
     scanf("%d",&cas);
     for(int t=;t<=cas;t++)
     {
         int n,m;
         scanf("%d%d",&n,&m);
         for(int i=;i<=id;i++) head[i]=dis1[i]=dis2[i]=;
         tot=;
         id=n;
         for(int i=;i<=m;i++)
         {
             int x,y;
             scanf("%d%d",&x,&y);
             id++;
             for(int j=;j<=y;j++)
             {
                 int z;
                 scanf("%d",&z);
                 add(z,id,x);
                 add(id,z,);
             }
         }
         for(int i=;i<=id;i++)
         {
             vis[i]=;
             dis1[i]=oo;
         }
         //for(int i=1;i<=id;i++) printf("%d %lld\n",i,dis1[i]);
         q.push(MP(,)); dis1[]=;
            while(!q.empty())
           {
                 int u=q.top().se;
                 q.pop();
                 if(vis[u]) continue;
              vis[u]=;
                 int e=head[u];
              while(e)
                 {
                     int v=vet[e];
                  if(dis1[u]+len[e]<dis1[v])
                     {
                         dis1[v]=dis1[u]+len[e];
                         q.push(MP(-dis1[v],v));
                    }
                     e=nxt[e];
                }
            }

            for(int i=;i<=id;i++)
         {
             vis[i]=;
             dis2[i]=oo;
         }
         q.push(MP(,n)); dis2[n]=;
            while(!q.empty())
           {
                 int u=q.top().se;
                 q.pop();
                 if(vis[u]) continue;
              vis[u]=;
                 int e=head[u];
              while(e)
                 {
                     int v=vet[e];
                  if(dis2[u]+len[e]<dis2[v])
                     {
                         dis2[v]=dis2[u]+len[e];
                         q.push(MP(-dis2[v],v));
                    }
                     e=nxt[e];
                }
            }
            //for(int i=1;i<=id;i++) printf("%d %lld %lld\n",i,dis1[i],dis2[i]);
            ll mn=oo;
            for(int i=;i<=n;i++) mn=min(mn,max(dis1[i],dis2[i]));
        //    printf("mn=%lld\n",mn);
         printf("Case #%d: ",t);
         if(mn==oo) printf("Evil John\n");
          else
          {
              int ans=;
              for(int i=;i<=n;i++)
               if(max(dis1[i],dis2[i])==mn) c[++ans]=i;
             printf("%lld\n",mn);
             for(int i=;i<=ans-;i++) printf("%d ",c[i]);
             printf("%d\n",c[ans]);
          }
     }
     return ;
 }