poj1778 All Discs Considered

思路：

拓扑排序。贪心。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 vector<int> G[];
 int n1, n2;
 inline void help(queue<int>& a, queue<int>& b, int x)
 {
     if (x > n1) b.push(x);
     else a.push(x);
 }
 bool same(int x, int y)
 {
     if (x >=  && x <= n1 && y >=  && y <= n1) return true;
     if (x > n1 && y > n1) return true;
     return false;
 }
 int cal(queue<int> a, queue<int> b, vector<int>in, bool set)
 {
     int cnt = , last = -, rnd = ;
     while (!a.empty() || !b.empty())
     {
         if ((!set && rnd) || set)
         {
             while (!a.empty() && !in[a.front()])
             {
                 int tmp = a.front(); a.pop();
                 if (!same(tmp, last)) cnt++;
                 last = tmp;
                 for (int i = ; i < (int)G[tmp].size(); i++)
                 {
                     int son = G[tmp][i];
                     if (--in[son] == ) help(a, b, son);
                 }
             }
         }
         while (!b.empty() && !in[b.front()])
         {
             int tmp = b.front(); b.pop();
             if (!same(tmp, last)) cnt++;
             last = tmp;
             for (int i = ; i < (int)G[tmp].size(); i++)
             {
                 int son = G[tmp][i];
                 if (--in[son] == ) help(a, b, son);
             }
         }
         rnd++;
     }
     return cnt;
 }
 int main()
 {
     int d, x, y;
     while (~scanf("%d %d %d", &n1, &n2, &d), n1 + n2 + d)
     {
         for (int i = ; i <= n1 + n2; i++) G[i].clear();
         vector<int> in(n1 + n2 + , );
         while (d--)
         {
             scanf("%d %d", &x, &y);
             G[y].push_back(x);
             in[x]++;
         }
         queue<int> a, b;
         for (int i = ; i <= n1 + n2; i++)
         {
             if (!in[i]) help(a, b, i);
         }
         printf("%d\n", min(cal(a, b, in, true), cal(a, b, in, false)) + );
     }
     return ;
 }