Title Address: POJ 1236
The main idea of this question is to ask at least how many points to send the message can make any one point can receive the message and the minimum number of additional edges can make the graph a connected graph. For the first problem, you can find the number of strong connected blocks with a degree of 0, because only a strong connected block with a degree of 0 is unable to receive information from the outside world, and as long as there is an entry, then the entire connected block can receive information. The second problem is the maximum of the number of strongly connected blocks with a degree of 0 and a strong connected block with a degree of 0.
The code is as follows:
#include <iostream> #include <cstdio> #include <string> #include <cstring> #include < stdlib.h> #include <math.h> #include <ctype.h> #include <queue> #include <map> #include < Set> #include <algorithm>using namespace Std;int head[200], CNT, index, top, ans, cntin, Cntout, in[200], out[200] int dfn[200], belong[200], instack[200], stak[200], low[200];struct node{int u, V, next;} edge[100000];void Add (int u , int v) {edge[cnt].v=v; Edge[cnt].next=head[u]; head[u]=cnt++;} void Init () {memset (head,-1,sizeof (head)); cnt=0; memset (instack,0,sizeof (instack)); memset (dfn,0,sizeof (DFN)); top=index=ans=0; cntin=cntout=0; memset (In,0,sizeof (in)); Memset (out,0,sizeof (out));} void Tarjan (int u) {dfn[u]=low[u]=++index; Instack[u]=1; Stak[++top]=u; for (int i=head[u]; i!=-1; i=edge[i].next) {int v=edge[i].v; if (!dfn[v]) {Tarjan (v); Low[u]=min (Low[v],low[u]); } else if (Instack[v]) low[u]=min (Low[u],dfn[v]); } if (Dfn[u]==low[u]) {ans++; while (1) {int v=stak[top--]; Belong[v]=ans; instack[v]=0; if (u==v) break; }}}int Main () {int n, I, J, A; scanf ("%d", &n); Init (); for (I=1; i<=n; i++) {while (scanf ("%d", &a) &&a) {Add (i,a); }} for (I=1; i<=n; i++) {if (!dfn[i]) Tarjan (i); } if (ans==1) printf ("1\n0\n"); else {for (i=1; i<=n; i++) {for (j=head[i]; j!=-1; j=edge[j].next) { int v=edge[j].v; if (Belong[i]!=belong[v]) {in[belong[v]]++; out[belong[i]]++; }}} for (I=1; i<=ans; i++) {if (!in[i]) cntin++; if (!out[i]) cntout++; } printf ("%d\n%d\n", Cntin,max (Cntin,cntout)); } return 0;}
POJ 1236 Network of schools (strong connected components)