UVALive - 5135 Mining Your Own Business

刘汝佳白书上面的一道题目：题意是给定一个联通分量，求出割顶以及双连通分量的个数，并且要求出安放安全井的种类数，也就是每个双连通分量中结点数（除开 割顶）个数相乘，对于有2个及以上割顶的双连通分量可以不用安放安全井。如果整个图就是一个双连通分量，那么需要安放两个安全井，种类数是n*(n-1)/2.

代码来自刘汝佳白书：

 #include <iostream>
 #include <sstream>
 #include <cstdio>
 #include <climits>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <stack>
 #include <map>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <algorithm>
 #define esp 1e-6
 #define pi acos(-1.0)
 #define pb push_back
 #define mp(a, b) make_pair((a), (b))
 #define in  freopen("in.txt", "r", stdin);
 #define out freopen("out.txt", "w", stdout);
 #define print(a) printf("%d\n",(a));
 #define bug puts("********))))))");
 #define stop  system("pause");
 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
 #define pragma comment(linker, "/STACK:102400000, 102400000")
 #define inf 0x0f0f0f0f

 using namespace std;
 typedef long long  LL;
 typedef vector<int> VI;
 typedef pair<int, int> pii;
 typedef vector<pii,int> VII;
 typedef vector<int>:: iterator IT;
 const int maxn =  ;
 struct Edge{
 int u, v;
 Edge(int u, int v):u(u), v(v){}
 };
 int pre[maxn], low[maxn], bccno[maxn], iscut[maxn], bcc_cnt, dfs_clock;
 VI g[maxn], bcc[maxn];
 stack<Edge> S;
 int dfs(int u, int fa)
 {
     int lowu = pre[u] = ++dfs_clock;
     int child = ;
     for(int i = ; i < g[u].size(); i++)
     {
         int v = g[u][i]; Edge e = Edge(u, v);
         if(!pre[v])
         {
             S.push(e);
             child++;
             int lowv = dfs(v, u);
             lowu = min(lowu, lowv);
             if(lowv >= pre[u])
             {
                 iscut[u] = ;
                 bcc_cnt++; bcc[bcc_cnt].clear();
                 for(;;)
                 {
                     Edge x = S.top(); S.pop();
                     if(bccno[x.u] != bcc_cnt) {bccno[x.u] = bcc_cnt; bcc[bcc_cnt].pb(x.u);}
                     if(bccno[x.v] != bcc_cnt) {bccno[x.v] = bcc_cnt; bcc[bcc_cnt].pb(x.v);}
                     if(x.u == u && x.v == v) break;
                 }
             }
         }
         else if(pre[v] < pre[u] && v!= fa)
         {
             S.push(e);
             lowu = min(lowu, pre[v]);
         }
     }
     if(child ==  && fa < ) iscut[u] = ;
     return low[u] = lowu;
 }
 void find_bcc(int n)
 {
     memset(iscut, , sizeof(iscut));
     memset(pre, , sizeof(pre));
     memset(bccno, , sizeof(bccno));

     dfs_clock = bcc_cnt = ;
     for(int i = ; i < n; i++)
         if(!pre[i]) dfs(i, -);
 }
 int kase;
 void solve(int n)
 {
     find_bcc(n);
     LL ans1 = , ans2 = ;
     for(int i = ; i <= bcc_cnt; i++)
     {
         int cut_cnt = ;
         for(int j = ; j < bcc[i].size(); j++)
             if(iscut[bcc[i][j]]) cut_cnt++;
         if(cut_cnt == )
             ans1++, ans2 *= (LL)(bcc[i].size() - cut_cnt);
     }
     if(bcc_cnt == )
     {
         ans1 = , ans2 = (LL)(n-)*n/;
     }
     printf("Case %d: %I64d %I64d\n", kase, ans1, ans2);
 }
 int main(void)
 {
     int m;
     while(scanf("%d", &m), m)
     {
         kase++;
         for(int i = ; i < maxn; i++)
             g[i].clear();
         int mxn = ;
         while(m--)
         {
             int u, v;
             scanf("%d%d", &u, &v);
             mxn = max(mxn, max(u, v));
             u--, v--;
             g[u].pb(v), g[v].pb(u);
         }
         solve(mxn);
     }
     return ;
 }

此题的另一种解法是先求出割顶，然后从非割顶的点dfs一遍，注意这个过程中不能经过割顶，统计每次dfs中遇到割顶的数目，如果为1，则更新答案。同样注意割顶为0的情况。

代码：

 #include <iostream>
 #include <sstream>
 #include <cstdio>
 #include <climits>
 #include <cstring>
 #include <cstdlib>
 #include <string>
 #include <stack>
 #include <map>
 #include <cmath>
 #include <vector>
 #include <queue>
 #include <algorithm>
 #define esp 1e-6
 #define pi acos(-1.0)
 #define pb push_back
 #define mp(a, b) make_pair((a), (b))
 #define in  freopen("in.txt", "r", stdin);
 #define out freopen("out.txt", "w", stdout);
 #define print(a) printf("%d\n",(a));
 #define bug puts("********))))))");
 #define stop  system("pause");
 #define Rep(i, c) for(__typeof(c.end()) i = c.begin(); i != c.end(); i++)
 #define inf 0x0f0f0f0f
 #pragma comment(linker, "/STACK:102400000,102400000")

 using namespace std;
 typedef long long  LL;
 typedef vector<int> VI;
 typedef pair<int, int> pii;
 typedef vector<pii,int> VII;
 typedef vector<int>:: iterator IT;
 const int maxn =  + ;
 int pre[maxn], low[maxn], iscut[maxn], dfs_clock;
 VI g[maxn], cut;
 int n;
 LL ans1, ans2, cut_cnt;
 int dfs(int u, int fa)
 {
     int lowu = pre[u] = ++dfs_clock;
     int child = ;
     for(int i = ; i < g[u].size(); i++)
     {
         int v = g[u][i];
         if(!pre[v])
         {
             child++;
             int lowv = dfs(v, u);
             lowu = min(lowu, lowv);
             if(lowv >= pre[u])
             {
                 iscut[u] = ;
             }
         }
         else if(pre[v] < pre[u] && v != fa)
         {
             lowu = min(lowu, pre[v]);
         }
     }
     if(child ==  && fa < )
         iscut[u] = ;
     return low[u] = lowu;
 }
 void find_bcc(int n)
 {
     memset(pre, , sizeof(pre));
     memset(iscut, , sizeof(iscut));
     dfs_clock = ;
     for(int  i = ; i < n; i++)
         if(!pre[i])
             dfs(i, -);
 }
 int cnt;
 void dfs1(int u)
 {
     cnt++;
     pre[u] = ;
     for(int i = ; i < g[u].size(); i++)
     {
         int v = g[u][i];
         if(!pre[v])
         {
             if(iscut[v]) cut_cnt++, cnt++, pre[v] = , cut.pb(v);
             else dfs1(v);
         }
     }
 }
 void solve(int n)
 {
     cut.clear();
     cut_cnt= ;
     memset(pre, , sizeof(pre));
     for(int i = ; i < n; i++)
         if(iscut[i]) cut_cnt++;

     if(cut_cnt == ) ans1 = , ans2 = (LL)n*(n-)/;
     else for(int i = ; i < n; i++)
             if(!pre[i] && !iscut[i])
             {
                 cut_cnt = cnt = ;
                 dfs1(i);
                 for(int i = ; i  < cut.size(); i++)
                     pre[cut[i]] = ;
                 cut.clear();
                 if(cut_cnt == )
                     ans1++, ans2 *= (LL)(cnt-);
             }
 }
 int main(void)
 {
     int m, t = ;
     while(scanf("%d", &m), m)
     {
         n = ;
         ans1 = , ans2 = ;
         for(int i = ; i < maxn; i++)
             g[i].clear();
         while(m--)
         {
             int u, v;
             scanf("%d%d", &u, &v);
             n = max(n, max(u, v));
             u--, v--;
             g[u].pb(v);
             g[v].pb(u);
         }
         find_bcc(n);
         solve(n);
         printf("Case %d: %I64d %I64d\n", t, ans1, ans2);
         t++;
     }
     return ;
 }