二分图判定+点染色/并查集 BestCoder Round #48 ($) 1002 wyh2000 and pupil

题目传送门

 /*
     二分图判定+点染色：因为有很多联通块，要对所有点二分图匹配，若不能，存在点是无法分配的，no
             每一次二分图匹配时，将点多的集合加大最后第一个集合去
     注意：n <= 1，no，两个集合都至少有一人；ans == n，扔一个给另一个集合
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <map>
 using namespace std;

 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 vector<int> G[MAXN];
 int col[MAXN];
 int cnt[];
 int n, m;

 bool DFS(int u)
 {
     for (int i=; i<G[u].size (); ++i)
     {
         int v = G[u][i];
         if (col[v] == -) {
             col[v] =  - col[u];    cnt[col[v]]++;
             if (!DFS (v))   return false;
         }
         else if (col[v] == col[u])  return false;
     }

     return true;
 }

 void work(void)
 {
     memset (col, -, sizeof (col));
     int ans = ;
     for (int i=; i<=n; ++i)
     {
         if (col[i] == -)
         {
             col[i] = ; cnt[] = ; cnt[] = ;
             if (!DFS (i)) {
                 puts ("Poor wyh");  return ;
             }
             ans += max (cnt[], cnt[]);
         }
     }
     if (ans == n)   ans--;
     printf ("%d %d\n", ans, n - ans);
 }

 int main(void)      //BestCoder Round #48 ($) 1002    wyh2000 and pupil
 {
     int t;   scanf ("%d", &t);
     while (t--)
     {
         scanf ("%d%d", &n, &m);
         if (n <= ) {
             puts ("Poor wyh");  continue;
         }
         for (int i=; i<=n; ++i)    G[i].clear ();
         for (int i=; i<=m; ++i)
         {
             int u, v;   scanf ("%d%d", &u, &v);
             G[u].push_back (v); G[v].push_back (u);
         }
         work ();
     }

     return ;
 }

 /*
     并查集：有点像POJ食物链的做法， 分为(1, n)和(n+1, 2*n)两个集合，两种情况都试一下。
             rt[i] == -1表示是该集合的根，用sz数组记录集合的大小
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 #include <string>
 #include <vector>
 #include <map>
 using namespace std;

 const int MAXN = 1e5 + ;
 const int INF = 0x3f3f3f3f;
 int n, m;
 struct UF   {
     int rt[MAXN*], sz[MAXN*], sz2[MAXN*];
     void init(void) {
         memset (rt, -, sizeof (rt));
         for (int i=; i<=*n; ++i)  sz[i] = (i <= n), sz2[i] = (i > n);
     }
     int Find(int x) {
         return rt[x] == - ? x : rt[x] = Find (rt[x]);
     }
     void Union(int x, int y)    {
         x = Find (x);   y = Find (y);
         if (x == y) return ;
         rt[x] = y;
         sz[y] += sz[x]; sz2[y] += sz2[x];
     }
     bool same(int x, int y) {
         return Find (x) == Find (y);
     }
 }uf;

 int main(void)      //BestCoder Round #48 ($) 1002    wyh2000 and pupil
 {
     int t;   scanf ("%d", &t);
     while (t--)
     {
         scanf ("%d%d", &n, &m);
         bool ok = true; uf.init ();
         for (int i=; i<=m; ++i)
         {
             int u, v;   scanf ("%d%d", &u, &v);
             if (!ok)   continue;
             if (uf.same (u, v) || uf.same (u+n, v+n))   ok = false;
             else    uf.Union (u, v+n), uf.Union (u+n, v);
         }

         if (!ok || n <= )    puts ("Poor wyh");
         else if (m == )    printf ("%d 1\n", n - );
         else    {
             int ans = ;
             for (int i=; i<=n; ++i)    {
                 if (uf.rt[i] == -) {
                     ans += min (uf.sz[i], uf.sz2[i]);
                 }
             }
             printf ("%d %d\n", n - ans, ans);
         }
     }

     return ;
 }

并查集做法