abc142_f

abc142_f

Description meaning of the questions:

Given a \ (n-\) points, \ (m \) have edges directed graph \ ((n \ le 1000, m \ le 2000) \)

Find a subgraph, such that subgraph in degree and out of each point is equal to \ (1 \)

That is to find a ring, the ring only point in an even outward side

Output program

Correct answer:

Hand play any discovery can first find a ring

Even if the inner edge of the ring, the ring put out, every point has been reduced to a degree within the ring are \ (1 \)

Condensed ring simulation process, as long as found in a loop to find the smallest ring can

Then use bfs seized a directed tree, the first atavistic edge that even the smallest ring two points where the answer is,

Find solutions, then you can remember what pre (precursor) in the path

Dfs also had a scene with, but is dqk hack out

Code：

/*
    bfs 找环
*/
#include <bits/stdc++.h>
#define N 1005
#define M 2005

using namespace std;

int n, m;
int dep[N];
int head[N], nex[M], to[M], e;
int stk[N], top;
bool instack[N], vis[N], G[N][N];
vector<int> res;

inline int Rd() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return x;
}
inline void Ae(int u, int v) {
    nex[++e] = head[u], head[u] = e;
    to[e] = v;
}

void calc(int u, int v) {
    do {
        res.push_back(stk[top]);
    } while(stk[top--] != v);
    int ans = res.size();
    printf("%d\n", ans);
    for(int i = 0; i < ans; ++i) printf("%d\n", res[i]);
    exit(0);
}
void dfs(int u) {
    stk[++top] = u;
    instack[u] = vis[u] = true;
    for(int i = head[u]; i; i = nex[i]) {
        int v = to[i];
        if(vis[v]) {
            if(instack[v]) calc(u, v);
        } else {
            dfs(v);
        }
    }
    instack[u] = false, --top;
    return;
}

int main() {
    n = Rd(), m = Rd();
    for(int i = 1; i <= m; ++i) {
        int u = Rd(), v = Rd();
        G[u][v] = true;
        Ae(u, v);
    }
    
    for(int i = 1; i <= n; ++i) {
        for(int j = 1; j <= n; ++j) {
            if(G[i][j] && G[j][i]) {
                puts("2");
                printf("%d\n%d\n", i, j);
                return 0;
            }   
        }
    }
    for(int i = 1; i <= n; ++i)
        if(!vis[i])
            dfs(i);
    puts("-1");
    return 0;   
}