poj3114 strong connectivity + shortest circuit

Test instructions: There are n cities, between cities can be sent by mail or e-mail messages, known m-mail lines, each line on behalf of A can send mail to B, and spend V time, if several cities can mail each other, then they are in the same country, they can send messages through e-mail, It takes 0 time. There are k queries, and each time you ask for a message from point A to B, the minimum amount of time it takes.

Since multiple cities can send mail to each other, it is not necessary to pass messages to each other in the same country, so it is first strongly connected and then the solution is done for each query. I started to think that strong connectivity after the indentation is directed to the non-circular graph, direct DFS for each query complexity is not very large, but the result of T, and then replaced by each query single source the shortest, then a dropped.

1#include <stdio.h>2#include <string.h>3#include <stack>4#include <queue>5#include <algorithm>6#include <vector>7 using namespacestd;8typedef pair<int,int>PII;9 Ten Const intmaxn=505; One Const intmaxm=250005; A Const intinf=0x3f3f3f3f; -  - inthead[2][maxn],point[2][maxm],nxt[2][maxm],size[2],val[2][MAXM]; the intn,t,scccnt; - intSTX[MAXN],LOW[MAXN],SCC[MAXN]; - intDIS[MAXN]; -stack<int>S; +  - intMinintAintb) {returnA<b?a:b;} +  A voidinit () { atmemset (head,-1,sizeof(head)); -size[0]=size[1]=0; - } -  - voidAddintAintBintVintC=0){ -point[c][size[c]]=b; innxt[c][size[c]]=Head[c][a]; -val[c][size[c]]=v; tohead[c][a]=size[c]++; + } -  the structcmp{//Change priority queue to small Gan *     BOOL operator() (PII a,pii b) { $         returnA.first>B.first;Panax Notoginseng     } - }; the  + voidDij (intSintT) {//incoming starting point and arrival point A     inti; thePriority_queue<pii,vector<pii>,cmp>Q; +Q.push (Make_pair (0, s)); -memset (DIS,0x3f,sizeof(DIS)); $dis[s]=0; $      while(!Q.empty ()) { -PII u=q.top (); - Q.pop (); the         if(U.first>dis[u.second])Continue; -          for(i=head[1][u.second];~i;i=nxt[1][i]) {Wuyi             intj=point[1][i]; the             if(dis[j]>u.first+val[1][i]) { -dis[j]=u.first+val[1][i]; Wu Q.push (Make_pair (dis[j],j)); -             } About         } $     } -     if(Dis[t]==inf) printf ("Nao e possivel entregar a carta\n"); -     Elseprintf"%d\n", Dis[t]); - } A  + voidDfsints) { thestx[s]=low[s]=++T; - S.push (S); $      for(inti=head[0][s];~i;i=nxt[0][i]) { the         intj=point[0][i]; the         if(!Stx[j]) { the Dfs (j); thelow[s]=min (low[s],low[j]); -         } in         Else if(!Scc[j]) { thelow[s]=min (low[s],stx[j]); the         } About     } the     if(low[s]==Stx[s]) { thescccnt++; the          while(1){ +             intu=S.top (); S.pop (); -scc[u]=scccnt; the             if(S==u) Break;Bayi         } the     } the } -  - voidSETSCC () { thememset (STX,0,sizeof(STX)); thememset (SCC,0,sizeof(SCC)); theT=scccnt=0; the      for(intI=1; i<=n;++i)if(!Stx[i]) DFS (i); -      for(intI=1; i<=n;++i) { the          for(intj=head[0][i];~j;j=nxt[0][j]) { the             intk=point[0][j]; the             if(scc[i]!=Scc[k]) {94Add (scc[i],scc[k],val[0][J],1); the             } the         } the     }98 } About  - intMain () {101     intm;102      while(SCANF ("%d", &n)!=eof&&N) {103scanf"%d",&m);104 init (); the          while(m--){106             inta,b,v;107scanf"%d%d%d",&a,&b,&v);108 Add (a,b,v);109         } the SETSCC ();111         intK; thescanf"%d",&k);113          while(k--){ the             intb; thescanf"%d%d",&a,&b); the             if(Scc[a]==scc[b]) printf ("0\n");117             Else{118 Dij (Scc[a],scc[b]);119             } -         }121printf"\ n");122     }123     return 0;124}

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poj3114 strong connectivity + shortest circuit