HDU 2242 双连通分量 考研路茫茫——空调教室

思路就是求边双连通分量，然后缩点，再用树形DP搞一下。

代码和求强连通很类似，有点神奇，=_=，慢慢消化吧

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <stack>
 using namespace std;

 const int maxn =  + ;
 const int maxm =  + ;

 int n, m;
 int a[maxn];

 struct Edge
 {
     int v, nxt;
 }edges[maxm * ];
 int ecnt;
 int head[maxn];

 bool vis[maxn];

 void AddEdge(int u, int v)
 {
     edges[ecnt].v = v;
     edges[ecnt].nxt = head[u];
     head[u] = ecnt++;
 }

 stack<int> S;
 int dfs_clock, scc_cnt;
 int low[maxn], pre[maxn], sccno[maxn], w[maxn];

 void dfs(int u, int fa)
 {
     low[u] = pre[u] = ++dfs_clock;
     S.push(u);

     for(int i = head[u]; ~i; i = edges[i].nxt)
     {
         if(i == (fa ^ )) continue;
         int v = edges[i].v;
         if(!pre[v])
         {
             dfs(v, i);
             low[u] = min(low[u], low[v]);
         }
         else if(!sccno[v]) low[u] = min(low[u], pre[v]);
     }

     if(pre[u] == low[u])
     {
         scc_cnt++;
         for(;;)
         {
             int x = S.top(); S.pop();
             sccno[x] = scc_cnt;
             w[scc_cnt] += a[x];
             if(x == u) break;
         }
     }
 }

 void find_scc()
 {
     scc_cnt = dfs_clock = ;
     memset(w, , sizeof(w));
     memset(pre, , sizeof(pre));
     memset(sccno, , sizeof(sccno));
     for(int i = ; i < n; i++) if(!pre[i]) dfs(i, -);
 }

 vector<int> G[maxn];

 void dfs2(int u)
 {
     vis[u] = true;
     for(int i = ; i < G[u].size(); i++)
     {
         int v = G[u][i];
         if(vis[v]) continue;
         dfs2(v);
         w[u] += w[v];
     }
 }

 int main()
 {
     while(scanf("%d%d", &n, &m) ==  && n)
     {
         int sum = ;
         for(int i = ; i < n; i++) { scanf("%d", a + i); sum += a[i]; }

         memset(head, -, sizeof(head));
         ecnt = ;

         while(m--)
         {
             int u, v; scanf("%d%d", &u, &v);
             AddEdge(u, v); AddEdge(v, u);
         }

         find_scc();

         if(scc_cnt == ) { puts("impossible"); continue; }

         for(int i = ; i <= scc_cnt; i++) G[i].clear();
         for(int u = ; u < n; u++)
             for(int i = head[u]; ~i; i = edges[i].nxt)
             {
                 int v = edges[i].v;
                 if(sccno[u] != sccno[v]) G[sccno[u]].push_back(sccno[v]);
             }

         memset(vis, false, sizeof(vis));
         dfs2();

         int ans = ;
         for(int i = ; i <= scc_cnt; i++) ans = min(ans, abs(sum - w[i] * ));
         printf("%d\n", ans);
     }

     return ;
 }

代码君