UVa 12167 & HDU 2767 强连通分量 Proving Equivalences
题意：给出一个有向图，问最少添加几条有向边使得原图强连通。
解法：求出SCC后缩点，统计一下出度为0的点和入度为0的点，二者取最大值就是答案。
还有个特殊情况就是本身就是强连通的话，答案就是0.
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <vector> 6 #include <stack> 7 using namespace std; 8 9 const int maxn = 20000 + 10; 10 11 int n, m; 12 13 vector<int> G[maxn]; 14 15 stack<int> S; 16 int pre[maxn], low[maxn], sccno[maxn]; 17 int dfs_clock, scc_cnt; 18 19 void dfs(int u, int fa) 20 { 21 low[u] = pre[u] = ++dfs_clock; 22 S.push(u); 23 24 for(int i = 0; i < G[u].size(); i++) 25 { 26 int v = G[u][i]; 27 if(!pre[v]) 28 { 29 dfs(v, u); 30 low[u] = min(low[u], low[v]); 31 } 32 else if(!sccno[v]) low[u] = min(low[u], pre[v]); 33 } 34 35 if(pre[u] == low[u]) 36 { 37 scc_cnt++; 38 for(;;) 39 { 40 int x = S.top(); S.pop(); 41 sccno[x] = scc_cnt; 42 if(x == u) break; 43 } 44 } 45 } 46 47 void find_scc() 48 { 49 memset(pre, 0, sizeof(pre)); 50 memset(sccno, 0, sizeof(sccno)); 51 dfs_clock = scc_cnt = 0; 52 for(int i = 1; i <= n; i++) if(!pre[i]) dfs(i, 0); 53 } 54 55 int in[maxn], out[maxn]; 56 57 int main() 58 { 59 int T; scanf("%d", &T); 60 while(T--) 61 { 62 scanf("%d%d", &n, &m); 63 for(int i = 1; i <= n; i++) G[i].clear(); 64 while(m--) 65 { 66 int u, v; scanf("%d%d", &u, &v); 67 G[u].push_back(v); 68 } 69 70 find_scc(); 71 72 if(scc_cnt == 1) { puts("0"); continue; } 73 74 memset(in, 0, sizeof(in)); 75 memset(out, 0, sizeof(out)); 76 for(int i = 1; i <= n; i++) 77 for(int j = 0; j < G[i].size(); j++) 78 { 79 int u = sccno[i], v = sccno[G[i][j]]; 80 if(u != v) { out[u]++; in[v]++; } 81 } 82 83 int hehe = 0, haha = 0; 84 for(int i = 1; i <= scc_cnt; i++) 85 { 86 if(!in[i]) hehe++; 87 if(!out[i]) haha++; 88 } 89 printf("%d\n", max(hehe, haha)); 90 } 91 92 return 0; 93 }
