Almost Acyclic Graph CodeForces - 915D （思维+拓扑排序判环）

Almost Acyclic Graph

CodeForces - 915D

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 1

output

Copy

YES

input

Copy

5 61 22 33 23 12 14 5

output

Copy

NO

Note

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.

题意：

给你有一个n个点，m个边的有向图。

问是否可以只删除一个边，使整个图无环。

思路：

枚举每一个节点，将该节点的入度减去1，先不用管删除的是哪个边，删除一个终点是i节点的边的影响就是i的入度减去1.

然后通过拓扑排序在\(O(n+m)\) 的时间复杂度里可以判断出一个有向图是否有环。

所以整体的时间复杂度是\(O(n*(n+m))\)

代码：

#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';
        }
    }
}

原文地址：https://www.cnblogs.com/qieqiemin/p/11620866.html