题目链接：http://poj.org/problem?id=1236
题意：
本题为有向图。
需解决两个问题：
1 须要给多少个点，才干传遍全部点。
2 加多少条边，使得整个图变得强连通。
使用Tarjan进行缩点，得到一个SCC图、
这个图有多少个入度为0的，多少个出度为0的。
如果有n个入度为0，m个出度为0
那么第一个答案就是n,第二个答案是max(n,m)
代码：
#include <stdio.h>
#include <iostream>
#include <math.h>
#include <stdlib.h>
#include <ctype.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <sstream>
#include <time.h>
using namespace std;
const int MAXN = 20010;
const int MAXM = 50010;
struct Edge
{
int to, next;
}edge[MAXM];
int head[MAXM], tot;
int Low[MAXN], Dfn[MAXN], Stack[MAXN], Belong[MAXN];//Belong的值为 1 ~ scc
int Index, top;
int scc;//强连通个数
bool Instack[MAXN];
int num[MAXN];//各个强连通包括的点的个数
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void Tarjan(int u)
{
int v;
Low[u] = Dfn[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
for (int i = head[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if (!Dfn[v])
{
Tarjan(v);
if (Low[u] > Low[v])
Low[u] = Low[v];
}
else if (Instack[v] && Low[u] > Dfn[v])
Low[u] = Dfn[v];
}
if (Low[u] == Dfn[u])
{
scc++;
do
{
v = Stack[--top];
Instack[v] = false;
Belong[v] = scc;
num[scc]++;
} while (v != u);
}
}
int in[MAXN], out[MAXN];
void solve(int N)
{
memset(Dfn,0,sizeof(Dfn));
memset(Instack,false,sizeof(Instack));
memset(num,0,sizeof(num));
Index = scc = top = 0;
for (int i = 1; i <= N; i++)
{
if (!Dfn[i])
Tarjan(i);
}
if (scc == 1)
{
printf("1\n0\n");
return;
}
for (int i = 1; i <= scc; i++)
in[i] = out[i] = 0;
for (int u = 1; u <= N; u++)
{
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (Belong[u] != Belong[v])
{
in[Belong[v]]++;
out[Belong[u]]++;
}
}
}
int ans1 = 0, ans2 = 0;
for (int i = 1; i <= scc; i++)
{
if (in[i] == 0) ans1++;
if (out[i] == 0) ans2++;
}
//printf("%d\n",scc);
printf("%d\n%d\n",ans1,max(ans1,ans2));
}
void init()
{
tot = 0;
memset(head,-1,sizeof(head));
}
int main()
{
int n;
int u, v;
while (~scanf("%d", &n))
{
init();
for (int i = 1; i <= n; i++)
{
while (~scanf("%d", &u) && u)
{
addedge(i,u);
}
}
solve(n);
}
return 0;
}