E. Quantifier Question （拓扑排序求前驱和后继）

题目链接：https://codeforces.com/contest/1345/problem/E

 

 

 

 

 

 

 

#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <queue>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <ctime>
#include <bitset>
#include <cmath>
#include <sstream>
#include <iostream>
#include <unordered_map>

#define ll long long
#define ull unsigned long long
#define ls nod<<1
#define rs (nod<<1)+1
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define INF 0x3f3f3f3f3f3f3f3f
#define max(a, b) (a>b?a:b)
#define min(a, b) (a<b?a:b)


const double eps = 1e-10;
const int maxn = 2e5 + 10;
const ll MOD = 99999999999999;

int sgn(double a) { return a < -eps ? -1 : a < eps ? 0 : 1; }

using namespace std;

vector<int> g[maxn],f[maxn];
int InDeg[maxn];
int pre[maxn],nxt[maxn];
bool ans[maxn];
int n,m;

bool Topsort(vector<int> &q) {
    q.clear();
    for (int i = 1;i <= n;i++) {
        if (!InDeg[i])
            q.push_back(i);
    }
    for (int i = 0;i < q.size();i++) {
        for (auto j : g[q[i]]) {
            if (--InDeg[j] == 0)
                q.push_back(j);
        }
    }
    return q.size() == n;
}
int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1;i <= m;i++) {
        int u,v;
        cin >> u >> v;
        g[u].push_back(v);
        f[v].push_back(u);
    }
    for (int i = 1;i <= n;i++) {
        InDeg[i] = f[i].size();
    }
    vector<int> top;
    if (!Topsort(top)) {
        cout << -1 << endl;
        return 0;
    }
    for (int i = 1;i <= n;i++) {
        pre[i] = nxt[i] = i;
    }
    // 找前驱
    for (int i = 0;i < top.size();i++) {
        for (int j : f[top[i]])
            pre[top[i]] = min(pre[top[i]],pre[j]);
    }
    // 找后继
    for (int i = top.size()-1;i >= 0;i--) {
        for (int j : g[top[i]]) {
            nxt[top[i]] = min(nxt[top[i]],nxt[j]);
        }
    }
    int cnt = 0;
    for (int i = 1;i <= n;i++) {
        if (min(pre[i],nxt[i]) == i) {
            cnt++;
            ans[i] = 1;
        }
    }
    cout << cnt << endl;
    for (int i = 1;i <= n;i++) {
        if (ans[i])
            cout << "A";
        else
            cout << "E";
    }
    cout << endl;
    return 0;
}