POJ 2942Knights of the Round Table(二分图判定+双连通分量)

Posted 2020-07-02 zhaop

题目链接

 

 

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 #include <algorithm>
  5 #include <stack>
  6 #include <vector>
  7 using namespace std;
  8 const int Max = 1010;
  9 vector<int> G[Max], bcc[Max];
 10 int odd[Max], color[Max];
 11 int A[Max][Max];
 12 int pre[Max], iscut[Max], bccno[Max];
 13 int dfs_clock, bcc_cnt;
 14 
 15 struct Edge
 16 {
 17     int u, v;
 18 };
 19 stack<Edge> S;
 20 int dfs(int u, int fa)
 21 {
 22     int lowu = pre[u] = ++dfs_clock;
 23     int child = 0;
 24     int Size = (int) G[u].size();
 25     for (int i = 0; i < Size; i++)
 26     {
 27         int v = G[u][i];
 28         Edge e;
 29         e.u = u;
 30         e.v = v;
 31         if (!pre[v])
 32         {
 33             S.push(e);
 34             child++;
 35             int lowv = dfs(v, u);
 36             lowu = min(lowu, lowv);
 37             if (lowv >= pre[u])
 38             {
 39                 iscut[u] = true;
 40                 ++bcc_cnt;
 41                 bcc[bcc_cnt].clear();
 42                 for(;;)
 43                 {
 44                     Edge x = S.top();
 45                     S.pop();
 46                     if (bccno[x.u] != bcc_cnt)
 47                     {
 48                         bcc[bcc_cnt].push_back(x.u);
 49                         bccno[x.u] = bcc_cnt;
 50                     }
 51                     if (bccno[x.v] != bcc_cnt)
 52                     {
 53                         bcc[bcc_cnt].push_back(x.v);
 54                         bccno[x.v] = bcc_cnt;
 55                     }
 56                     if (x.u == u && x.v == v)
 57                         break;
 58                 }
 59             }
 60         }
 61         else if (pre[v] < pre[u] && v != fa)
 62         {
 63             S.push(e);
 64             lowu = min(lowu, pre[v]);
 65         }
 66     }
 67     if (fa < 0 && child == 1)
 68         iscut[u] = false;
 69     return lowu;
 70 }
 71 void find_bcc(int n)
 72 {
 73     memset(pre, 0, sizeof(pre));
 74     memset(iscut, 0, sizeof(iscut));
 75     memset(bccno, 0, sizeof(bccno));
 76     dfs_clock = bcc_cnt = 0;
 77     for (int i = 0; i < n; i++)
 78     {
 79         if (!pre[i])
 80             dfs(i, -1);
 81     }
 82 }
 83 bool bipartite(int u, int b)
 84 {
 85     for (int i = 0; i < (int) G[u].size(); i++)
 86     {
 87         int v = G[u][i];
 88         if (bccno[v] != b)
 89             continue;
 90         if (color[v] == color[u])
 91             return false;
 92         if (!color[v])
 93         {
 94             color[v] = 3 - color[u];
 95             if (!bipartite(v, b))
 96                 return false;
 97         }
 98     }
 99     return true;
100 }
101 int main()
102 {
103     int n, m;
104     while (scanf("%d%d", &n, &m) != EOF)
105     {
106         if (n == 0 && m == 0)
107             break;
108         for (int i = 0; i <= n; i++)
109             G[i].clear();
110         memset(A, 0, sizeof(A));
111         for (int i = 0; i < m; i++)
112         {
113             int u, v;
114             scanf("%d%d", &u, &v);
115             u--;
116             v--;
117             A[u][v] = A[v][u] = 1;
118         }
119         for (int i = 0; i < n; i++)
120         {
121             for (int j = i + 1; j < n; j++)
122             {
123                 if (!A[i][j])
124                 {
125                     G[i].push_back(j);
126                     G[j].push_back(i);
127                 }
128             }
129         }
130 
131         find_bcc(n);
132 
133        // cout << bcc[3][0] << endl;
134         memset(odd, 0, sizeof(odd));
135         for (int i = 1; i <= bcc_cnt; i++)
136         {
137             memset(color, 0, sizeof(color));
138             for (int j = 0; j < (int) bcc[i].size(); j++)
139                 bccno[ bcc[i][j] ] = i;
140             int u = bcc[i][0];
141             color[u] = 1;
142 
143             if (!bipartite(u, i))
144             {
145                 for (int j = 0; j < (int) bcc[i].size(); j++)
146                     odd[ bcc[i][j] ] = 1;
147             }
148         }
149         int ans = n;
150         for (int i = 0; i < n; i++)
151             if (odd[i])
152             ans--;
153         printf("%d\n", ans);
154     }
155     return 0;
156 }

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