【Floyd(并非水题orz)】BZOJ4093-[Usaco2013 Dec]Vacation Planning

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 const int MAXK=200+5;
  5 const int MAXN=20000+50;
  6 const ll INF=1e12;
  7 struct node
  8 {
  9     int to,dis;
 10 };
 11 vector<node> E[MAXN];
 12 int n,m,k,q;
 13 ll d[MAXK][MAXK],dis[MAXK][MAXN];
 14 int id[MAXN],num[MAXN];
 15 
 16 void addedge(int u,int v,int w)
 17 {
 18     E[u].push_back((node){v,w}); 
 19 }
 20 
 21 void init()
 22 {
 23     scanf("%d%d%d%d",&n,&m,&k,&q);
 24     memset(id,0,sizeof(id));
 25 
 26     for (int i=1;i<=m;i++)
 27     {
 28         int u,v,w;
 29         scanf("%d%d%d",&u,&v,&w);
 30         addedge(u,v,w);
 31     }
 32     for (int i=1;i<=k;i++)
 33     {
 34         int x;
 35         scanf("%d",&x);
 36         id[x]=i,num[i]=x;
 37     }
 38 }
 39 
 40 void prep()
 41 {        
 42     for (int i=1;i<=k;i++)
 43         for (int j=1;j<=k;j++)
 44             d[i][j]=INF;
 45     for (int i=1;i<=k;i++) d[i][i]=0;
 46     for (int i=1;i<=k;i++)
 47     {
 48         int u=num[i];
 49         for (int j=E[u].size()-1;j>=0;j--)
 50         {
 51             int v=E[u][j].to;
 52             if (id[v]) d[i][id[v]]=min(d[i][id[v]],(ll)E[u][j].dis);
 53                 else
 54                 {
 55                     for (int _k=E[v].size()-1;_k>=0;_k--)
 56                     {
 57                         int vto=E[v][_k].to;
 58                         if (id[vto] && vto!=num[i]) d[id[u]][id[vto]]=min(d[id[u]][id[vto]],(ll)E[u][j].dis+(ll)E[v][_k].dis);
 59                     }
 60                 }
 61         }
 62     }    
 63     
 64     for (int _k=1;_k<=k;_k++)
 65         for (int i=1;i<=k;i++)
 66             for (int j=1;j<=k;j++)
 67                 if (i!=j && j!=_k && _k!=i) d[i][j]=min(d[i][j],d[i][_k]+d[_k][j]); 
 68 }
 69 
 70 void prep2()
 71 {
 72     for (int i=1;i<=k;i++)
 73         for (int j=1;j<=n;j++) dis[i][j]=INF;
 74     for (int i=1;i<=k;i++)
 75         for (int j=1;j<=k;j++) dis[i][num[j]]=d[i][j];
 76         
 77         
 78         
 79     for (int i=1;i<=k;i++)
 80     {
 81         for (int _k=0;_k<E[num[i]].size();_k++)//注意这里是E[num[i]]不是num[i]，检查了40分钟才发现QAQ
 82             for (int j=1;j<=k;j++)
 83             {
 84                 int to=E[num[i]][_k].to;
 85                 dis[j][to]=min(dis[j][to],d[j][i]+E[num[i]][_k].dis);
 86             }
 87     }
 88 }
 89 
 90 void solve()
 91 {
 92     int t=0;
 93     ll totalans=0;
 94     for (int i=0;i<q;i++)
 95     {
 96         int a,b;
 97         scanf("%d%d",&a,&b);
 98         ll ans=INF;
 99         if (id[a]) ans=dis[id[a]][b];
100             else
101             {
102                 for (int j=0;j<E[a].size();j++)
103                 {
104                     int v=E[a][j].to;
105                     if (id[v]) ans=min(ans,E[a][j].dis+dis[id[v]][b]);
106                 }
107             }
108         if (ans<INF)
109         {
110             totalans+=ans;
111             t++;
112         }
113     }
114     printf("%d\n",t);
115     printf("%d",totalans);
116 }
117 
118 int main()
119 {
120     init();
121     prep();
122     prep2();
123     solve();
124     return 0;
125 }
126 
127 /*
128 TLE的solve，答案是正确的QAQ 
129 void solve()
130 {
131     int t=0;
132     ll totalans=0;
133     for (int i=1;i<=q;i++)
134     {
135         int a,b;
136         ll ans=INF;
137         scanf("%d%d",&a,&b);
138         if (id[a] && id[b]) ans=d[id[a]][id[b]];
139         else if (id[a])
140         {
141             for (int ib=0;ib<rE[b].size();ib++) ans=min(ans,d[id[a]][id[rE[b][ib].to]]+(ll)rE[b][ib].dis);
142         }
143         else if (id[b])
144         {
145             for (int ia=0;ia<E[a].size();ia++) ans=min(ans,d[id[E[a][ia].to]][id[b]]+(ll)E[a][ia].dis);
146         }
147         else
148         {
149             for (int ia=0;ia<E[a].size();ia++)
150                 for (int ib=0;ib<rE[b].size();ib++)
151                 {
152                     ll now=(ll)E[a][ia].dis+(ll)rE[b][ib].dis;
153                     now+=d[id[E[a][ia].to]][id[rE[b][ib].to]];
154                     ans=min(ans,now);
155                 }
156         }
157         if (ans!=INF) t++,totalans+=ans;
158     }
159     printf("%d\n",t);
160     printf("%lld",totalans);
161 }*/