题目大意：给出N个点，M条边，要求你添加最少的边，使得这个图变成强连通分量

解题思路：先找出所有的强连通分量和桥，将强连通分量缩点，桥作为连线，就形成了DAG了

这题被坑了。用了G＋＋交的，结果一直RE，用C＋＋一发就过了。。。

#include 
   
     #include 
    
      #define N 20010 #define M 100010 #define min(a,b) ((a) > (b)? (b): (a)) #define max(a,b) ((a) > (b)? (a): (b)) struct Edge{ int from, to, next; }E[M]; int head[N], sccno[N], pre[N], lowlink[N], stack[N], in[N], out[N]; int n, m, tot, dfs_clock, top, scc_cnt; void AddEdge(int from, int to) { E[tot].from = from; E[tot].to = to; E[tot].next = head[from]; head[from] = tot++; } void init() { memset(head, -1, sizeof(head)); tot = 0; int u, v; for (int i = 0; i < m; i++) { scanf(%d%d, &u, &v); AddEdge(u, v); } } void dfs(int u) { pre[u] = lowlink[u] = ++dfs_clock; stack[++top] = u; for (int i = head[u]; i != -1; i = E[i].next) { int v = E[i].to; if (!pre[v]) { dfs(v); lowlink[u] = min(lowlink[u], lowlink[v]); } else if (!sccno[v]) { lowlink[u] = min(lowlink[u], pre[v]); } } int x; if (pre[u] == lowlink[u]) { scc_cnt++; while (1) { x = stack[top--]; sccno[x] = scc_cnt; if (x == u) break; } } } void solve() { memset(sccno, 0, sizeof(sccno)); memset(pre, 0, sizeof(pre)); dfs_clock = top = scc_cnt = 0; for (int i = 1; i <= n; i++) if (!pre[i]) dfs(i); if (scc_cnt <= 1) { printf(0 ); return ; } for (int i = 1; i <= scc_cnt; i++) in[i] = out[i] = 1; for (int i = 0; i < tot; i++) { int u = E[i].from, v = E[i].to; if (sccno[u] != sccno[v]) { out[sccno[u]] = in[sccno[v]] = 0; } } int a = 0, b = 0; for (int i = 1; i <= scc_cnt; i++) { if (out[i]) a++; if (in[i]) b++; } printf(%d , max(a, b)); } int main() { while (scanf(%d%d, &n, &m) != EOF) { init(); solve(); } return 0; } 
    
   

?