Catowice City【Codeforces 1248 F】【Tarjan】

Codeforces Round #594 (Div. 2) F

  这道题的解法还真是不少，写了个枚举也可以做这道题，当然Tarjan自然也是可以的。

  我一开始没捋清楚思路，再想想，发现，我们看到审判者，他们都会指向一些参赛选手，那么我们是不是可以尽力去找那些没有指向的人，也就是出度为0的点，那么他们岂不是就没有指向了。然后我们可以把有关系的缩点，然后再DAG上找到一个审判就可以了，剩下的都是参赛者了。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define eps 1e-9
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e6 + 7;
int N, M, head[maxN], cnt;
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
int dfn[maxN], low[maxN], tot, Stop, Stap[maxN], Belong[maxN], Bcnt, du[maxN], siz[maxN];
bool instack[maxN];
void Tarjan(int u)
{
    dfn[u] = low[u] = ++tot;
    Stap[++Stop] = u;
    instack[u] = true;
    for(int i=head[u], v; ~i; i=edge[i].nex)
    {
        v = edge[i].to;
        if(!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v]) low[u] = min(low[u], dfn[v]);
    }
    int v;
    if(low[u] == dfn[u])
    {
        Bcnt++; siz[Bcnt] = 0;
        do
        {
            v = Stap[Stop--];
            instack[v] = false;
            Belong[v] = Bcnt;
            siz[Bcnt]++;
        } while(u != v);
    }
}
inline void init()
{
    cnt = tot = Stop = Bcnt = 0;
    for(int i=1; i<=N; i++)
    {
        head[i] = -1;
        du[i] = dfn[i] = 0;
        instack[i] = false;
    }
}
int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &N, &M);
        init();
        for(int i=1, u, v; i<=M; i++)
        {
            scanf("%d%d", &u, &v);
            if(u == v) continue;
            addEddge(u, v);
        }
        for(int i=1; i<=N; i++) if(!dfn[i]) Tarjan(i);
        if(Bcnt == 1) { printf("No\n"); continue; }
        for(int u=1; u<=N; u++)
        {
            for(int i=head[u], v; ~i; i=edge[i].nex)
            {
                v = edge[i].to;
                if(Belong[u] == Belong[v]) continue;
                du[Belong[u]]++;
            }
        }
        int id;
        for(id = 1; id <= Bcnt; id++) if(!du[id]) break;
        printf("Yes\n");
        printf("%d %d\n", siz[id], N - siz[id]);
        for(int i=1; i<=N; i++) if(Belong[i] == id) printf("%d ", i); puts("");
        for(int i=1; i<=N; i++) if(Belong[i] ^ id) printf("%d ", i); puts("");
    }
    return 0;
}