【并查集】 Codeforces Round #286 B - Mr. Kitayuta's Technology

对每一个联通块做并查集，然后每一个联通块里最多只有一个环，然后只要找有多少个环就行啦。。。

#include <iostream>#include <queue> #include <stack> #include <map> #include <set> #include <bitset> #include <cstdio> #include <algorithm> #include <cstring> #include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 200005#define maxm 400005#define eps 1e-10#define mod 1000000007#define INF 0x3f3f3f3f#define PI (acos(-1.0))#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid #define rson o<<1 | 1, mid+1, R#define pii pair<int, int>//#pragma comment(linker, "/STACK:16777216")typedef long long LL;typedef unsigned long long ULL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}// headstruct node{int v;node *next;}*H[maxn], E[maxm], *edges;bool vis1[maxn];bool vis2[maxn];bool cyc[maxn];int f[maxn];int n, m, ans;void addedges(int u, int v){edges->v = v;edges->next = H[u];H[u] = edges++;}int find(int u){return f[u] = u == f[u] ? f[u] : find(f[u]);}bool merge(int a, int b){int aa = find(a), bb = find(b);if(aa == bb) return false;f[aa] = bb;return true;}void init(void){edges = E;ans = 0;memset(H, 0, sizeof H);}void read(void){int u, v;scanf("%d%d", &n, &m);for(int i = 0; i <= n; i++) f[i] = i;while(m--) {scanf("%d%d", &u, &v);addedges(u, v);ans += merge(u, v);}}void dfs(int u){vis1[u] = 1;for(node *e = H[u]; e; e = e->next) {if(vis2[e->v]) cyc[find(u)] = true;if(vis1[e->v]) continue;vis2[e->v] = true;dfs(e->v);vis2[e->v] = false;}}void work(void){for(int i = 1; i <= n; i++) if(!vis1[i]) {vis2[i] = true;dfs(i);vis2[i] = false;}for(int i = 1; i <= n; i++) ans += cyc[i];printf("%d\n", ans);}int main(void){init();read();work();return 0;}