1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐

1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 432  Solved: 270
[Submit][Status]

Description

The cows are having a picnic! Each of Farmer John's K (1 <= K <= 100) cows is grazing in one of N (1 <= N <= 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 <= M <= 10,000) one-way paths (no path connects a pasture to itself). The cows want to gather in the same pasture for their picnic, but (because of the one-way paths) some cows may only be able to get to some pastures. Help the cows out by figuring out how many pastures are reachable by all cows, and hence are possible picnic locations.

  K(1≤K≤100)只奶牛分散在N(1≤N≤1000)个牧场．现在她们要集中起来进餐．牧场之间有M(1≤M≤10000)条有向路连接，而且不存在起点和终点相同的有向路．她们进餐的地点必须是所有奶牛都可到达的地方．那么，有多少这样的牧场呢？

Input

* Line 1: Three space-separated integers, respectively: K, N, and M * Lines 2..K+1: Line i+1 contains a single integer (1..N) which is the number of the pasture in which cow i is grazing. * Lines K+2..M+K+1: Each line contains two space-separated integers, respectively A and B (both 1..N and A != B), representing a one-way path from pasture A to pasture B.

第1行输入K，N，M.接下来K行，每行一个整数表示一只奶牛所在的牧场编号．接下来M行，每行两个整数，表示一条有向路的起点和终点

Output

* Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.

所有奶牛都可到达的牧场个数

Sample Input

2 4 4
2
3
1 2
1 4
2 3
3 4

INPUT DETAILS:

4<--3
^ ^
| |
| |
1-->2

The pastures are laid out as shown above, with cows in pastures 2 and 3.

Sample Output

2

牧场3，4是这样的牧场．

HINT

 

Source

Silver

题解：尼玛这道题居然都被卡了一次——原因很逗比，因为中间BFS当此点访问过时应该跳过，结果我一开始脑抽写了个 if c[p^.g]=1 then continue; 仔细想想，当这种情况下p指针还没等跳到下一个就continue了啊，不死循环才怪！！！（phile:多大了还犯这种错！！！）。。。然后没别的了，就是对于每个牛都用BFS或者DFS来搜一编能够到达的点，然后没了——复杂度才O(K(N+M))肯定没问题。。。（所以一开始当我看到红色的TLE时真心被吓到了QAQ）

 type
     point=^node;
     node=record
                g:longint;
                next:point;
     end;

 var
    i,j,k,l,m,n,f,r:longint;
    P:point;
    a:array[..] of point;
    b,c,d,e:array[..] of longint;
 procedure add(x,y:longint);inline;
           var p:point;
           begin
                new(p);
                p^.g:=y;
                p^.next:=a[x];
                a[x]:=p;
           end;
 begin
      readln(e[],n,m);
      for i:= to n do
          begin
               d[i]:=;
               a[i]:=nil;
          end;
      for i:= to e[] do
               readln(e[i]);
      for i:= to m do
          begin
               readln(j,k);
               add(j,k);
          end;
      for i:= to e[] do
          begin
               fillchar(c,sizeof(c),);
               fillchar(b,sizeof(b),);
               c[e[i]]:=;
               b[]:=e[i];
               f:=;r:=;
               while f<r do
                     begin
                          p:=a[b[f]];
                          while p<>nil do
                                begin
                                     if c[p^.g]= then
                                        begin
                                             c[p^.g]:=;
                                             b[r]:=p^.g;
                                             inc(r);
                                        end;
                                     p:=p^.next;
                                end;
                          inc(f);
                     end;
               for j:= to n do
                   d[j]:=d[j]*c[j];
          end;
      l:=;
      for i:= to n do l:=l+d[i];
      writeln(l);
 end.