codeforces 659E . New Reform 强连通

题目链接

对于每一个联通块， 如果有一个强连通分量， 那么这个联通块对答案的贡献就是0。 否则对答案贡献是1.

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 1e5+5;
int num, head[maxn], s[maxn], ok[maxn], dfn[maxn], low[maxn];
int instack[maxn], st[maxn], vis[maxn], deep, cnt, top;
struct node
{
    int to, nextt;
}e[maxn*2];
void add(int u, int v) {
    e[num].to = v, e[num].nextt = head[u], head[u] = num++;
}
void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++deep;
    st[++top] = u;
    instack[u] = 1;
    for(int i = head[u]; ~i; i = e[i].nextt) {
        int v = e[i].to;
        if(v == fa)
            continue;
        if(!dfn[v]) {
            tarjan(v, u);
            low[u] = min(low[u], low[v]);
        } else if(instack[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(dfn[u] == low[u]) {
        ++cnt;
        int v;
        do {
            v = st[top--];
            instack[v] = 0;
            s[v] = cnt;
        } while(u != v);
    }
}
int bfs(int u) {
    queue <int> q;
    q.push(u);
    vis[u] = 1;
    int flag = 0;
    while(!q.empty()) {
        u = q.front(); q.pop();
        if(ok[u])
            flag = 1;
        for(int i = head[u]; ~i ; i = e[i].nextt) {
            int v = e[i].to;
            if(vis[v])
                continue;
            q.push(v);
            vis[v] = 1;
        }
    }
    return !flag;
}
pll ed[maxn];
int main()
{
    int n, m;
    cin>>n>>m;
    mem1(head);
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &ed[i].fi, &ed[i].se);
        add(ed[i].fi, ed[i].se);
        add(ed[i].se, ed[i].fi);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i])
            tarjan(i, 0);
    num = 0;
    mem1(head);
    for(int i = 0; i < m; i++) {
        int u = s[ed[i].fi], v = s[ed[i].se];
        if(u == v) {
            ok[u] = 1;
            continue;
        }
        add(u, v);
        add(v, u);
    }
    int ans = 0;
    for(int i = 1; i <= cnt; i++) {
        if(!vis[i]) {
            ans += bfs(i);
        }
    }
    cout<<ans<<endl;
    return 0;
}