【IOI 1996】 Network of Schools

【题目链接】

         点击打开链接

【算法】

          对于第一问，将这个图缩点，输出出度为零的点的个数

          对于第二问，同样将这个图缩点，输出入度为零、出度为零的点的个数的最大值

【代码】

          

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 10010

struct Edge
{
		int to,nxt;
} e[MAXN];

int i,x,n,tot,timer,top,cnt;
int head[MAXN],dfn[MAXN],low[MAXN],stk[MAXN],belong[MAXN];
bool instack[MAXN];

template <typename T> inline void read(T &x)
{
    int f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x/10);
    putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}
inline void add(int u,int v)
{
		tot++;
		e[tot] = (Edge){v,head[u]};
		head[u] = tot;			
}
inline void tarjan(int u)
{
		int i,v;
		dfn[u] = low[u] = ++timer;
		instack[u] = true;
		stk[++top] = u;
		for (i = head[u]; i; i = e[i].nxt)
		{
				v = e[i].to;		
				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])
		{
				cnt++;
				while (stk[top+1] != u)
				{
						instack[stk[top]] = false;
						belong[stk[top]] = cnt;
						top--;
				}
		}
}
inline void solve()
{
		int i,j,s1 = 0,s2 = 0;
		static int in[MAXN],out[MAXN];
		memset(in,0,sizeof(in));
		memset(out,0,sizeof(out));
		memset(dfn,0,sizeof(dfn));
		memset(low,0,sizeof(low));
		cnt = timer = 0;
		for (i = 1; i <= n; i++)
		{
				if (!dfn[i])
						tarjan(i);	
		}	
		for (i = 1; i <= n; i++)
		{
				for (j = head[i]; j; j = e[j].nxt)
				{
						if (belong[i] != belong[e[j].to])
						{
								in[belong[e[j].to]]++;
								out[belong[i]]++; 
						}
				}
		}
		for (i = 1; i <= cnt; i++)
		{
				if (!in[i]) s1++;
				if (!out[i]) s2++;
		}
		if (cnt == 1) printf("%d\n%d\n",1,0);
		else printf("%d\n%d\n",s1,max(s1,s2));
}

int main() {
		
		while (scanf("%d",&n) != EOF)
		{
				tot = 0;
				for (i = 1; i <= n; i++) head[i] = 0;
				for (i = 1; i <= n; i++)
				{
						while (scanf("%d",&x) && x) add(i,x);	
				}	
				solve();
		}
		
		return 0;
	
}