这题怎么这么水~~~本来以为挺难的一道题，结果随便一写就过了。。。本来还不知道损坏的牛棚算不算，结果不明不白就过了。。。
题干：
农夫John的农场遭受了一场地震.有一些牛棚遭到了损坏,但幸运地,所有牛棚间的路经都还能使用. FJ的农场有P(1 <= P <= 30,000)个牛棚,编号1..P. C(1 <= C <= 100,000)条双向路经联接这些牛棚,编号为1..C. 路经i连接牛棚a_i和b_i (1 <= a_i<= P;1 <= b_i <= P).路经可能连接a_i到它自己,两个牛棚之间可能有多条路经.农庄在编号为1的牛棚. N (1 <= N <= P)头在不同牛棚的牛通过手机短信report_j(2 <= report_j <= P)告诉FJ它们的牛棚(report_j)没有损坏,但是它们无法通过路经和没有损坏的牛棚回到到农场. 当FJ接到所有短信之后,找出最小的不可能回到农庄的牛棚数目.这个数目包括损坏的牛棚. 注意:前50次提交将提供在一些测试数据上的运行结果. 输入输出格式 输入格式： * Line 1: Three space-separated integers: P, C, and N * Lines 2..C+1: Line i+1 describes cowpath i with two integers: a_i and b_i * Lines C+2..C+N+1: Line C+1+j contains a single integer: report_j 输出格式： * Line 1: A single integer that is the minimum count of pastures from which a cow can not return to the barn (including the damaged pastures themselves) 输入输出样例 输入样例#1： 复制 4 3 1 1 2 2 3 3 4 3 输出样例#1： 复制 3
代码：
#include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<algorithm> #include<cstring> using namespace std; #define duke(i,a,n) for(int i = a;i <= n;i++) #define lv(i,a,n) for(int i = a;i >= n;i--) #define clean(a) memset(a,0,sizeof(a)) const int INF = 1 << 30; typedef long long ll; typedef double db; template <class T> void read(T &x) { char c; bool op = 0; while(c = getchar(), c < '0' || c > '9') if(c == '-') op = 1; x = c - '0'; while(c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0'; if(op) x = -x; } template <class T> void write(T x) { if(x < 0) putchar('-'), x = -x; if(x >= 10) write(x / 10); putchar('0' + x % 10); } int p,c,n,lst[60005],len = 0; struct node { int l,r,nxt; }a[200005]; void add(int x,int y) { a[++len].l = x; a[len].r = y; a[len].nxt = lst[x]; lst[x] = len; // cout<<len<<" "<<a[len].l<<" "<<a[len].r<<endl; } int vis[60005]; void dfs(int x) { if(vis[x] == -1 || vis[x] == 1) return; vis[x] = 1; for(int k = lst[x];k;k = a[k].nxt) { int y = a[k].r; if(vis[y] == 0) dfs(y); } } int main() { clean(vis); read(p);read(c);read(n); duke(i,1,c) { int g,h; read(g); read(h); add(g,h); add(h,g); } duke(i,1,n) { int k; read(k); vis[k] = -1; for(int j = lst[k];j;j = a[j].nxt) { int y = a[j].r; vis[y] = -1; } } dfs(1); int tot = 0; duke(i,1,p) { if(vis[i] == 0 || vis[i] == -1) tot++; } write(tot); return 0; }