题目大意：给一个有向图，问至少几条边，使其成为强连通图。
首先强联通缩点，然后分别统计入度为0的点数num1和出度为0的点数num2，答案就是max(num1, num2)。
为什么呢？不难想，入度为0说明没有点能到达他，所以必须连一条通向他的边；出度为0说明他不能通向任何点，所以也得连一条出边。于是这些点可以互相连，多出去的就再连别的点。
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 puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 2e4 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n, m; 38 vector<int> v[maxn]; 39 40 stack<int> st; 41 bool in[maxn]; 42 int dfn[maxn], low[maxn], cnt = 0; 43 int col[maxn], ccol = 0; 44 void tarjan(int now) 45 { 46 dfn[now] = low[now] = ++cnt; 47 st.push(now); in[now] = 1; 48 for(int i = 0; i <(int)v[now].size(); ++i) 49 { 50 if(!dfn[v[now][i]]) 51 { 52 tarjan(v[now][i]); 53 low[now] = min(low[now], low[v[now][i]]); 54 } 55 else if(in[v[now][i]]) low[now] = min(low[now], dfn[v[now][i]]); 56 } 57 if(dfn[now] == low[now]) 58 { 59 int x; ++ccol; 60 do 61 { 62 x = st.top(); st.pop(); 63 in[x] = 0; 64 col[x] = ccol; 65 }while(x != now); 66 } 67 } 68 69 int ind[maxn], oud[maxn], Mi = 0, Mo = 0; 70 void newGraph(int now) 71 { 72 int u = col[now]; 73 for(int i = 0; i < (int)v[now].size(); ++i) 74 { 75 int e = col[v[now][i]]; 76 if(u == e) continue; 77 oud[u]++; ind[e]++; 78 } 79 } 80 81 void init() 82 { 83 for(int i = 0; i < maxn; ++i) v[i].clear(); 84 while(!st.empty()) st.pop(); 85 Mem(dfn, 0); Mem(low, 0); Mem(in, 0); 86 Mem(col, 0); 87 cnt = ccol = 0; 88 Mem(ind, 0); Mem(oud, 0); Mi = Mo = 0; 89 } 90 91 int main() 92 { 93 int T = read(); 94 while(T--) 95 { 96 init(); 97 n = read(); m = read(); 98 for(int i = 1; i <= m; ++i) 99 { 100 int x = read(), y = read(); 101 v[x].push_back(y); 102 } 103 for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i); 104 for(int i = 1; i <= n; ++i) newGraph(i); 105 if(ccol == 1) {write(0), enter; continue;} 106 for(int i = 1; i <= ccol; ++i) 107 { 108 if(!ind[i]) Mi++; 109 if(!oud[i]) Mo++; 110 } 111 write(max(Mi, Mo)); enter; 112 } 113 return 0; 114 }