[USACO09JAN]Earthquake Damage

嘟嘟嘟

刚开始因为没看到只能走没有损坏的农场，磨叽了20多分钟……不管了，写题解吧。

首先如果一个点不能到达原点，那么和他相邻的点也不能到达原点，所以刚开始我们把不能走的点和他相邻的点都打上标记，然后跑dfs就行了。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<vector>
 8 #include<queue>
 9 #include<stack>
10 #include<cctype>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(' ')
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const db eps = 1e-8;
19 const int maxn = 3e4 + 5;
20 inline ll read()
21 {
22     ll ans = 0;
23     char ch = getchar(), last = ' ';
24     while(!isdigit(ch)) last = ch, ch = getchar();
25     while(isdigit(ch)) ans = (ans << 3) + (ans << 1) + ch - '0', ch = getchar();
26     if(last == '-') ans = -ans;
27     return ans;
28 }
29 inline void write(ll x)
30 { 
31     if(x < 0) putchar('-'), x = -x;
32     if(x >= 10) write(x / 10);
33     putchar(x % 10 + '0');
34 }
35 
36 int p, c, n;
37 vector<int> v[maxn];
38 bool vis[maxn];
39 int ans;
40 
41 void dfs(int now)
42 {
43     vis[now] = 1; ans--;
44     for(int i = 0; i < (int)v[now].size(); ++i)
45         if(!vis[v[now][i]]) dfs(v[now][i]);
46 }
47 
48 int main()
49 {
50     p = read(); c = read(); n = read();
51     for(int i = 1; i <= c; ++i)
52     {
53         int x = read(), y = read();
54         v[x].push_back(y); v[y].push_back(x);
55     }
56     for(int i = 1; i <= n; ++i)
57     {
58         int x = read(); vis[x] = 1;
59         for(int j = 0; j < (int)v[x].size(); ++j) vis[v[x][j]] = 1;
60     }
61     ans = p; dfs(1);
62     write(ans); enter;
63     return 0;
64 }

标签：now,ch,int,Damage,read,USACO09JAN,ans,include,Earthquake
来源： https://blog.51cto.com/u_15234622/2830904