告诉你一些点不能到达1,由于是双向边,也就是1不能到达那些点
那么从1开始dfs,如果当前点能到达不能到达的点,那么当前点就是损坏的。
#include <cstdio> #include <cstring> #include <iostream> #define N 100001 int n, m, p, cnt, tot; int head[N], to[N << 1], next[N << 1]; bool vis[N], flag[N]; inline int read() { int x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; return x * f; } inline void add(int x, int y) { to[cnt] = y; next[cnt] = head[x]; head[x] = cnt++; } inline void dfs(int u) { int i, v; vis[u] = 1; for(i = head[u]; i ^ -1; i = next[i]) { v = to[i]; if(flag[v]) return; } for(i = head[u]; i ^ -1; i = next[i]) { v = to[i]; if(!vis[v]) dfs(v); } tot++; } int main() { int i, x, y; n = read(); m = read(); p = read(); memset(head, -1, sizeof(head)); for(i = 1; i <= m; i++) { x = read(); y = read(); add(x, y); add(y, x); } for(i = 1; i <= p; i++) { x = read(); flag[x] = 1; } dfs(1); printf("%d ", n - tot); return 0; }