Almost Acyclic Graph
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a directed graph consisting of n vertices and m edges (each edge is directed, so it can be traversed in only one direction). You are allowed to remove at most one edge from it.
Can you make this graph acyclic by removing at most one edge from it? A directed graph is called acyclic iff it doesn't contain any cycle (a non-empty path that starts and ends in the same vertex).
Input
The first line contains two integers n and m (2 ≤ n ≤ 500, 1 ≤ m ≤ min(n(n - 1), 100000)) — the number of vertices and the number of edges, respectively.
Then m lines follow. Each line contains two integers u and v denoting a directed edge going from vertex u to vertex v (1 ≤ u, v ≤ n, u ≠ v). Each ordered pair (u, v) is listed at most once (there is at most one directed edge from u to v).
Output
If it is possible to make this graph acyclic by removing at most one edge, print YES. Otherwise, print NO.
Examples
input
Copy
3 41 22 33 23 1output
Copy
YESinput
Copy
5 61 22 33 23 12 14 5output
Copy
NONote
In the first example you can remove edge 
, and the graph becomes acyclic.
In the second example you have to remove at least two edges (for example, 
and ) in order to make the graph acyclic.
Meaning of the questions:
You have a point n, m have the sides of FIG.
Q. Can I delete just one side, so that the whole map acyclic.
Ideas:
Enumerate each node, the node into the minus 1 degree, not control which side is deleted, delete an end point is the impact side of the i-node is minus 1 degree i.
Then topological sort \ (O (n + m) \) time complexity where there can be judged whether there is a circumferential FIG.
Therefore, the overall time complexity is \ (O (n * (n + m)) \)
Code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline void getInt(int *p);
// const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
const int maxn = 510;
const int maxm = 3e5 + 10;
struct edge {
int to, from, nxt;
} edges[maxm];
int n, ind[maxn];
int in[maxn];
int head[maxn], cnt;
// 初始化
void init(int _n)
{
n = _n, cnt = -1;
for (int i = 1; i <= n; i++) { head[i] = -1, ind[i] = 0; }
}
// 加边
void addedge(int u, int v)
{
edges[++cnt].from = u;
edges[cnt].to = v;
edges[cnt].nxt = head[u];
head[u] = cnt;
ind[v]++;
}
bool go()
{
queue<int> Q;
for (int i = 1; i <= n; i++) {
if (ind[i] == 0) { Q.push(i); }
}
cnt = 0;
while (!Q.empty()) {
int u = Q.front();
Q.pop();
cnt++;
for (int i = head[u]; i != -1; i = edges[i].nxt) {
int v = edges[i].to;
if (--ind[v] == 0) { Q.push(v); }
}
}
return cnt == n;
}
int m;
int x, y;
int main()
{
//freopen("D:\\code\\text\\input.txt","r",stdin);
//freopen("D:\\code\\text\\output.txt","w",stdout);
du2(n, m);
init(n);
while (m--) {
du2(x, y);
addedge(x, y);
in[y]++;
}
int isok = 0;
repd(i, 1, n) {
if (in[y]) {
memcpy(ind, in, sizeof(in));
ind[i]--;
if (go()) {
isok = 1;
break;
}
}
}
if (isok) {
puts("YES");
} else {
puts("NO");
}
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}