【IOI 1996】 Network of Schools
【题目链接】
【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; }