BZOJ1718 [Usaco2006 Jan] Redundant Paths 分离的路径

给你一个无向图，问至少加几条边可以使整个图变成一个双联通分量

简单图论练习= =

先缩点，ans = (度数为1的点的个数) / 2

这不是很好想的么QAQ

然后注意位运算的优先级啊魂淡！！！你个sb调了一个下午！！！

 /**************************************************************
     Problem: 1718
     User: rausen
     Language: C++
     Result: Accepted
     Time:44 ms
     Memory:3148 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 const int N = 5e4 + ;
 const int M = 1e5 + ;

 struct edge {
     int next, to;
     edge(int _n = , int _t = ) : next(_n), to(_t) {}
 } e[M];

 int n, m, ans;
 int cnt, top, num, tot = ;
 int dfn[N], low[N], vis[N], sz[N], s[N], w[N], first[N];
 int mp[N];

 inline void Add_Edges(int x, int y) {
     e[++tot] = edge(first[x], y), first[x] = tot;
     e[++tot] = edge(first[y], x), first[y] = tot;
 }

 void DFS(int p, int from) {
     dfn[p] = low[p] = ++cnt;
     s[++top] = p, vis[p] = ;
 #define y e[x].to
     register int x;
     for (x = first[p]; x; x = e[x].next)
         if (x != (from ^ )) {
             if (!vis[y]) DFS(y, x);
             if (vis[y] < ) low[p] = min(low[p], low[y]);
         }
 #undef y
     if (dfn[p] == low[p]) {
         register int y;
         ++num;
         while (s[top + ] != p) {
             y = s[top--];
             vis[y] = , w[y] = num;
             ++sz[num];
         }
     }
 }

 inline void tarjan() {
     int i;
     cnt = top = num = ;
     memset(vis, sizeof(vis), );
     for (i = ; i <= n; ++i)
         if (!vis[i]) DFS(i, );
 }

 int main() {
     int i, x, y;
     scanf("%d%d", &n, &m);
     for (i = ; i <= m; ++i) {
         scanf("%d%d", &x, &y);
         Add_Edges(x, y);
     }
     tarjan();
 #define y e[x].to
     memset(mp, , sizeof(mp));
     for (i = ; i <= n; ++i)
         for (x = first[i]; x; x = e[x].next)
             if (w[i] != w[y]) ++mp[w[y]];
     for (ans = i = ; i <= num; ++i)
         ans += (mp[i] == );
     printf("%d\n", ans >> );
 #undef y
     return ;