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

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <vector>
  5 #include <stack>
  6 using namespace std;
  7 
  8 const int maxn = 10000 + 10;
  9 const int maxm = 20000 + 10;
 10 
 11 int n, m;
 12 int a[maxn];
 13 
 14 struct Edge
 15 {
 16     int v, nxt;
 17 }edges[maxm * 2];
 18 int ecnt;
 19 int head[maxn];
 20 
 21 bool vis[maxn];
 22 
 23 void AddEdge(int u, int v)
 24 {
 25     edges[ecnt].v = v;
 26     edges[ecnt].nxt = head[u];
 27     head[u] = ecnt++;
 28 }
 29 
 30 stack<int> S;
 31 int dfs_clock, scc_cnt;
 32 int low[maxn], pre[maxn], sccno[maxn], w[maxn];
 33 
 34 void dfs(int u, int fa)
 35 {
 36     low[u] = pre[u] = ++dfs_clock;
 37     S.push(u);
 38 
 39     for(int i = head[u]; ~i; i = edges[i].nxt)
 40     {
 41         if(i == (fa ^ 1)) continue;
 42         int v = edges[i].v;
 43         if(!pre[v])
 44         {
 45             dfs(v, i);
 46             low[u] = min(low[u], low[v]);
 47         }
 48         else if(!sccno[v]) low[u] = min(low[u], pre[v]);
 49     }
 50 
 51     if(pre[u] == low[u])
 52     {
 53         scc_cnt++;
 54         for(;;)
 55         {
 56             int x = S.top(); S.pop();
 57             sccno[x] = scc_cnt;
 58             w[scc_cnt] += a[x];
 59             if(x == u) break;
 60         }
 61     }
 62 }
 63 
 64 void find_scc()
 65 {
 66     scc_cnt = dfs_clock = 0;
 67     memset(w, 0, sizeof(w));
 68     memset(pre, 0, sizeof(pre));
 69     memset(sccno, 0, sizeof(sccno));
 70     for(int i = 0; i < n; i++) if(!pre[i]) dfs(i, -1);
 71 }
 72 
 73 vector<int> G[maxn];
 74 
 75 void dfs2(int u)
 76 {
 77     vis[u] = true;
 78     for(int i = 0; i < G[u].size(); i++)
 79     {
 80         int v = G[u][i];
 81         if(vis[v]) continue;
 82         dfs2(v);
 83         w[u] += w[v];
 84     }
 85 }
 86 
 87 int main()
 88 {
 89     while(scanf("%d%d", &n, &m) == 2 && n)
 90     {
 91         int sum = 0;
 92         for(int i = 0; i < n; i++) { scanf("%d", a + i); sum += a[i]; }
 93 
 94         memset(head, -1, sizeof(head));
 95         ecnt = 0;
 96 
 97         while(m--)
 98         {
 99             int u, v; scanf("%d%d", &u, &v);
100             AddEdge(u, v); AddEdge(v, u);
101         }
102 
103         find_scc();
104 
105         if(scc_cnt == 1) { puts("impossible"); continue; }
106 
107         for(int i = 1; i <= scc_cnt; i++) G[i].clear();
108         for(int u = 0; u < n; u++)
109             for(int i = head[u]; ~i; i = edges[i].nxt)
110             {
111                 int v = edges[i].v;
112                 if(sccno[u] != sccno[v]) G[sccno[u]].push_back(sccno[v]);
113             }
114 
115         memset(vis, false, sizeof(vis));
116         dfs2(1);
117 
118         int ans = 1000000000;
119         for(int i = 1; i <= scc_cnt; i++) ans = min(ans, abs(sum - w[i] * 2));
120         printf("%d\n", ans);
121     }
122 
123     return 0;
124 }
代码君

 

posted @ 2015-08-11 17:02  AOQNRMGYXLMV  阅读(207)  评论(0)  编辑  收藏  举报