POJ 1236 Network of schools

Main topic:

Given a direction graph, ask:
1) At least a few vertices must be selected in order to get from these vertices
All vertices can be reached.
2) At least how many edges must be added to make the top
Point of departure, all the vertices can be reached
Number of vertices <= 100

A useful theorem:

A point in which all the degrees in the non-circular graph are not 0. Must be from a certain degree of 0 points from the point of no-ring, so from any of the points not 0 to go back must be terminated at a point with a degree of 0

Problem Solving Ideas:

1. Find all the strongly connected components

2. Each strongly connected component is shrunk to a point, forming a directed acyclic graph Dag.

3. What is the number of vertices with a 0 in the Dag above, and the answer to question 1?

To add a few edges on a dag to make the Dag strong, the answer to question 2 is how much

The method of adding edges: To add an edge for each point with a degree of 0, to add an edge for each point with a degree of 0, to assume that there are N degrees 0 points, m out of a 0 point, Max (M,n) is the solution of the second problem (proving difficult, slightly)

But better think, I want to let those into the degree of 0, and the degree of 0, the point becomes not 0, so that we can form a complete unicom, we have a penetration of 0 of the connection a degree of 0, can disappear two pairs, the rest, we each

Just add one more side to the bar.

When we find the strong Unicom components can be, and then re-composition, and finally arrive at the result, there is an answer to special judgment, because there is only one strong unicom component when there is no need for extra edges, so to special judgment.

#include <iostream>#include<cstdlib>#include<cstdio>#include<algorithm>#include<vector>#include<queue>#include<cmath>#include<stack>#include<cstring>UsingNamespace std;#defineINF 0XFFFFFFF#defineMAXN 105#defineMin (A, b) (A&LT;B?A:B)BOOLINSTACK[MAXN];intLOW[MAXN], DFN[MAXN], STACK[MAXN], top, time, CNT;intMAPS[MAXN][MAXN];intBELONG[MAXN];intN;vector<int>G[MAXN];voidInit () {memset (Low,0,sizeof(low)); memset (DFN,0,sizeof(DFN)); memset (Instack,false,sizeof(Instack)); memset (Maps,0,sizeof(maps)); CNT= time = top =0;  for(intI=0; i<=n; i++) G[i].clear ();}voidTarjan (intu) {Low[u]= Dfn[u] = + +Time ; Stack[top++] =u; Instack[u]=true; intLen =g[u].size (), V;  for(intI=0; i<len; i++) {v=G[u][i]; if( !Low[v])            {Tarjan (v); Low[u]=min (Low[u], low[v]); } elseif (Instack[v]) {Low[u]=min (Low[u], dfn[v]); }    }    if(Low[u] = =Dfn[u]) {         Do{v= stack[--top]; BELONG[V]=CNT; INSTACK[V]=false; } while(U! =v); CNT++; }}voidsolve () {intINDEGREE[MAXN] = {0}; intOUTDEGREE[MAXN] = {0}; intin =0, out =0;  for(intI=1; i<=n; i++)    {        if(!Low[i])    Tarjan (i); }     for(intI=1; i<=n; i++)    {        intLen =g[i].size (), V;  for(intj=0; j<len; J + +) {v=G[i][j]; Maps[belong[i]][belong[v ]=1; }    }     for(intI=0; i<cnt; i++)    {         for(intj=0; j<cnt; J + +)        {            if(i = =j)Continue; if(Maps[i][j]) {Indegree[j]++; Outdegree[i]++; }        }    }     for(intI=0; i<cnt; i++)    {        if(Indegree[i] = =0) in++; if(Outdegree[i] = =0) out++; } printf ("%d\n", in); if(CNT = =1) printf ("0\n"); Elseprintf ("%d\n", Max (in, Out));}intMain () { while(SCANF ("%d", &n)! =EOF)        {Init ();  for(intI=1; i<=n; i++)        {            intA;  while(SCANF ("%d",&a), a) g[i].push_back (a);    } solve (); } Return0;}

POJ 1236 Network of schools