"Strong connected components" relay _insert

1, the Messenger

(Message.pas/c/cpp)

"Problem description"

Students from the interest group from various schools, in order to increase the friendship, the party has a word of the game, if a know B, then a received a message, it will pass this message to B, and all the people who know.

If a knows b,b not necessarily know a.

All the people from 1 to n numbers, give all "understanding" relationship, ask if I release a message, then will pass through a number of messages, this message back to I,1<=i<=n.

"Input File"

The first line in the input file message.in is two numbers n (n<1000) and M (m<10000), with a space between two numbers indicating the number and the cognitive relationship.

The next m line, two numbers a and b per line, represents a knowing B. 1<=a,b<=n. The cognitive relationship may be repeated, but a line of two numbers will not be the same.

"Output File"

Output file Message.out A total of n rows, one character T or F per line. Line I if it is T, I send a message back to I; if it is F, I send a message that does not pass back to I.

"Input Sample"

4 6

1 2

2 3

4 1

3 1

1 3

2 3

"Output Sample"

T

T

T

F

In fact, this problem is only using a DFS, for each point search once, see whether can search the starting point, O (n^2) algorithm can also pass.

I used a tarjan, if the strong connected component is greater than 1, the ring, O (n) is indicated.

Relatively fixed, or very simple.

#include <algorithm> #include <cstdio> using Std::min;
	struct Node {long ind;
Node* NXT;
};

node* head[1010];
	Long Getint () {long rs=0;bool Sgn=1;char tmp;
	do tmp = GetChar ();
	while (!isdigit (TMP) &&tmp-'-');
	if (tmp== '-') {Tmp=getchar (); sgn=0;}
	Do rs= (rs<<1) + (rs<<3) +tmp-' 0 ';
	while (IsDigit (Tmp=getchar ()));
return sgn?rs:-rs;
	} void Insert (long A,long b) {node* NN = new node;
	NN-> ind = b;
	nn-> nxt = head[a];
Head[a] = nn;
Long time = 0;
Long bcnt = 0;
Long belong[1010];
Long dfn[1010];
Long low[1010];
BOOL instack[1010];
Long stack[1010];
BOOL hash[1010][1010];
Long cnt[1010];

Long top = 0;
	void Tarjan (Long u) {dfn[u] = low[u] = ++time;
	Stack[++top] = u;
	Instack[u] = true;
		for (node* vv=head[u];vv;vv=vv->nxt) {Long v = vv->ind; if (!
			Dfn[v]) {Tarjan (v);
		Low[u] = min (low[u],low[v]);
		else if (Instack[v]) {Low[u] = min (low[u],dfn[v));
		} if (dfn[u] = = Low[u]) {bcnt + +;
		Long V; Do {v = stack[top--];
			INSTACK[V] = false;
			BELONG[V] = bcnt;
		CNT[BCNT] + +;
	}while (U!= v);
	int main () {freopen ("message.in", "R", stdin);
	Freopen ("Message.out", "w", stdout);
	Long n = getint ();
	Long m = Getint ();
		for (long i=1;i<m+1;i++) {Long a = Getint ();
		Long B = Getint ();
			if (!hash[a][b]) {hash[a][b] = true;
		Insert (A,B); for (long i=1;i<n+1;i++) if (!
	Dfn[i]) Tarjan (i);
		for (long i=1;i<n+1;i++) {if (Cnt[belong[i]] > 1) printf ("t\n");
	else printf ("f\n");
return 0;
 }