[HAOI2006]受欢迎的牛
考虑建图:如果A喜欢B,那么从A 到B就有一条边。有因为牛之间的喜爱关系有传递性,所以如果途中存在一个环的话,那就说明这个环中的任意一头牛都会被喜爱。那么自然可以想到用tarjan缩点来简化图。这样在一个DAG中,会发现被该联通块喜爱的牛一定是没有出边的。因为这张图可能不连通,所以还有记录这个出边为0的点的个数,如果大于1,就说明图不连通,就输出0,否则输出这个点的大小。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter printf("\n") 13 #define space printf(" ") 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const int eps = 1e-8; 19 const int maxn = 1e4 + 5; 20 inline ll read() 21 { 22 ll ans = 0; 23 char ch = getchar(), last = ' '; 24 while(!isdigit(ch)) {last = ch; ch = getchar();} 25 while(isdigit(ch)) 26 { 27 ans = ans * 10 + ch - '0'; ch = getchar(); 28 } 29 if(last == '-') ans = -ans; 30 return ans; 31 } 32 inline void write(ll x) 33 { 34 if(x < 0) x = -x, putchar('-'); 35 if(x >= 10) write(x / 10); 36 putchar(x % 10 + '0'); 37 } 38 39 int n, m; 40 vector<int> v[maxn]; 41 42 int dfn[maxn], low[maxn], cnt = 0; 43 bool in[maxn]; 44 stack<int> st; 45 int col[maxn], ccol = 0, val[maxn]; 46 void tarjan(int now) 47 { 48 dfn[now] = low[now] = ++cnt; 49 st.push(now); in[now] = 1; 50 for(int i = 0 ; i < (int)v[now].size(); ++i) 51 { 52 if(!dfn[v[now][i]]) 53 { 54 tarjan(v[now][i]); 55 low[now] = min(low[now], low[v[now][i]]); 56 } 57 else if(in[v[now][i]]) low[now] = min(low[now], dfn[v[now][i]]); 58 } 59 if(low[now] == dfn[now]) 60 { 61 int x; ccol++; 62 do 63 { 64 x = st.top(); 65 in[x] = 0; st.pop(); 66 col[x] = ccol; val[ccol]++; 67 }while(x != now); 68 } 69 return; 70 } 71 72 int out[maxn]; 73 74 int main() 75 { 76 n = read(); m = read(); 77 for(int i = 1; i <= m; ++i) 78 { 79 int x = read(), y = read(); 80 v[x].push_back(y); 81 } 82 for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i); 83 for(int i = 1; i <= n; ++i) 84 for(int j = 0; j < (int)v[i].size(); ++j) 85 if(col[i] != col[v[i][j]]) out[col[i]]++; 86 int tot = 0, pos; 87 for(int i = 1; i <= ccol; ++i) 88 { 89 if(!out[i]) tot++, pos = i; 90 if(tot > 1) {printf("0\n"); return 0;} 91 } 92 write(val[pos]); enter; 93 return 0; 94 }