Poj2186popular cows "Strong Connect" "Debug Two Hours"

I've been debugging for two hours, and I've been crying.

Write a little bit of experience

To tell you that a herd of cows has n-head (N<10^4)

Then tell you m to envy relationship such as a bull envy B bull

And envy can pass like a envy b b envy C then a envy C

Now ask how many cows are admired by all the cows

Analysis: My first thought was to traverse each cow and then build the opposite side from each cow and then DFS to see if it could sweep all the cows, but the time complexity O (n) cannot withstand

Later thought can be like flody delivery closure, but the same time complexity withstand

And then I thought this could be strong connectivity.

The first thing we want is a question.
For a direction-free graph

Satisfy an ox to be envied by all cows is: 1 figure connected 2 The cow's out of 0 (because he is the envy of all the cows and then if he envied the other cows will form a ring)

For the subject is a ring with strong connectivity to shrink point into a direction-free graph and then according to the two conditions to determine the

But the first thing I wrote was that I had isolated points in the picture.

Then it was after the break of the brain. It hurts to think.

Code:

1#include <iostream>2#include <cstdio>3#include <cstring>4 using namespacestd;5 6 Const intMAXN =50005;7 8 intN, M;9 inttot, CNT;Ten structNode { One     intto, next; A }E[MAXN]; - intHEAD[MAXN]; -  the voidAddintUintv) { -E[tot].to =v; -E[tot].next =Head[u]; -Head[u] = tot++; + } - intS[MAXN]; + intINS[MAXN]; A intSC; at intans; - intNUM[MAXN]; -  - intLOW[MAXN], DFN[MAXN]; -  - voidinit () { intot =1; -Memset (Head,0,sizeof(head)); tomemset (Low,0,sizeof(Low)); +memset (DFN,0,sizeof(DFN)); -memset (INS,0,sizeof(INS)); thememset (NUM,0,sizeof(num)); *CNT =1; $sc =0;Panax NotoginsengAns =0; - } the  + voidDfsintu) { AS[sc++] =u; theIns[u] =1; +      for(inti = Head[u]; I i =E[i].next) { -         intv =e[i].to; $         if(!Dfn[v]) { $DFN[V] = low[v] = cnt++; - Dfs (v); -Low[u] =min (Low[u], low[v]); the}Else if(Ins[v] &&!Num[v]) { -Low[u] =min (Low[u], dfn[v]);Wuyi         } the     } -     if(Low[u] = =Dfn[u]) { Wuans++; -         intx; About          Do { $x = s[--SC]; -NUM[X] =ans; -INS[X] =0; -} while(X! =u); A     } +  the } - intOUTED[MAXN]; $  the intMain () { the     intu, v; the      while(EOF! = scanf ("%d%d", &n, &m)) { the init (); -          for(inti =1; I <= m; i++) { inscanf"%d%d", &u, &v); the Add (U, v); the         } About          for(inti =1; I <= N; i++) { the             if(!Dfn[i]) { theDfn[i] = low[i] = cnt++; the DFS (i); +             } -         } the     //for (int i = 1; I <= n; i++) {Bayi     //printf ("%d", Num[i]); the     //} puts (""); the         intAns_num =-1; -memset (outed,0,sizeof(outed)); -          for(inti =1; I <= N; i++) { the //printf ("i =%d\n", i); the             if(Num[i] = =0) { theAns_num =0; the                  Break; -             } the              for(intj = Head[i]; J; j =E[j].next) { the                 intx =e[j].to; the                 if(Num[i]! =Num[x]) {94outed[num[i]]++; the                 } the             } the         }98     //for (int i = 1; I <= n; i++) { About     //printf ("*%d", Outed[i]); -     //} puts ("");101     //printf ("ans =%d\n", ans);102         intFlag =0;103         if(Ans_num! =0) {104Ans_num =0; the             intAx =0;106              for(inti =1; I <= ans; i++) {107                 if(Outed[i] = =0) {108flag++;109                     if(Flag >=2) { theAns_num =0;111                          Break; the                     }113Ax =i; the                 } the             } the             if(Flag <2) {117                  for(inti =1; I <= N; i++) {118                     if(Num[i] = =Ax) {119ans_num++; -                     }121                 }122             }123         }124printf"%d\n", ans_num); the     }126     return 0;127}

View Code

Poj2186popular cows "Strong Connect" "Debug Two Hours"