Poj1094-sorting It all out-topology sequencing

Test Instructions:

1). Give you some capital letters, a total of n; there is m-bar relationship between capital letters;

2). Example: Relationship: a<b, meaning that A is in front of B (that is, maybe B is in front of a)

3). Output: There are three kinds of cases:

1.N letters under the relationship of the former I can determine the sort relationship;

2.N Letters An error occurred when executing to the article I command, i.e. the loop: a indirect or direct less than B and b indirectly or directly less than A;

The 3.M command cannot determine the N-letter ordering.

Ideas:

The intuitive feeling is the topological sort, it should be noted that each read into a command to sort the topology once.

If you don't know enough about topological sequencing, see this blog: Click to view

and the order of three cases is more important!!

conditions 1 and 2 can be judged at the same time, finally judging the situation 3!!

It means:

If the execution of an order to article I can be determined in case 1 or 2, then the direct output;

If the M command is finished and the alphabetical order is not determined, the output condition 3

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AC Code:

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<Set>#include<string>#include<stack>#include<cmath>#defineTest printf ("***\n")#defineMM1 (a) memset ((a), -1,sizeof ((a)))#defineMM0 (a) memset ((a), 0,sizeof ((a)))#defineMMX (a) memset ((a), 0x3f,sizeof ((a)))#defineKa GetChar (); GetChar ()#defineKA1 GetChar ()#defineIIS Std::ios::sync_with_stdio (FALSE)using namespacestd; typedefLong LongLL; typedef unsignedLong LongULL; Const intN = the; Const intM =4100005; Const intX =999983; Const intINF =1e9; Const DoubleEPS = 1e-8; Const intMoD = 1e9 +7; //#define DEBUGintn,m,sum; intTot,flag; int inch[N],head[n],tin[n]; intAr[n]; structLP {intU,v,nex; LP () {} LP (intAintBintc): U (A), V (b), NEX (c) {}}cw[n*N]; InlinevoidAddintAintb) {cw[++tot]=LP (A,b,head[a]); Head[a]=tot; }  intTuopu () {Queue<int>Q;  while(!Q.empty ())      Q.pop ();  for(intI=0; i<=n;++i) tin[i]=inch[i];  for(intI=1; i<=n;++i) {          if(tin[i]==0) {Q.push (i); }      }      intt=0; intans=0;  while(!Q.empty ()) {          if(Q.size () >1) ans=-1; intu=Q.front ();          Q.pop (); Ar[t++]=u;  for(inti=head[u];i!=-1; i=Cw[i].nex) {              intv=cw[i].v; TIN[V]--; if(tin[v]==0) {Q.push (v); }          }      }      if(t!=n)return 1; returnans; } Inlinevoidinit () {tot=-1; MM0 (inch);  MM1 (head); }  intMainintargcChar Const*argv[]) {#ifdef DEBUG freopen ("d:in.in","R", stdin); Freopen ("D:out.out","W", stdout); #endif      Charb;  while(~SCANF ("%d%d", &n,&m) && (n+m))          {init (); intans=1;  for(intI=0; i<m;++i) {              Chars[5]; scanf ("%s", s); A=s[0];b=s[2]; if(ans==0)Continue; inch[B-'A'+1]++; Add (a-'A'+1, B-'A'+1); intok=Tuopu (); if(ok==1) {ans=0; printf ("inconsistency found after%d relations.\n", i+1); }Else if(ok==0) {ans=0; printf ("Sorted sequence determined after%d relations:", i+1);  for(intI=0; i<n;++i) {printf ("%c", ar[i]+'A'-1); } printf (". \ n"); }          }          if(ans) {printf ("Sorted sequence cannot be determined.\n"); }      }      return 0; }  

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Poj1094-sorting It all out-topology sequencing