【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;
    
}