传送门:CodeForces-1217D
思路分析
如果没有环答案肯定是1
如果有环答案一定是2
因为有向图的环中,肯定存在由大指向小和由小指向大两种边
所以只需要将这两种边染不同色就行了
样例输入
3 3
1 2
2 3
3 1
样例输出
2
1 1 2
AC代码
#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <chrono>
#include <random>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) x.size()
#define lowbit(x) x&(-x)
#define all(x) x.begin(),x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int INF = 0x3f3f3f3f;
const int N=1e5+10;
int a[N],b[N],in[N],n,m;
vector<int>e[N];
bool tp() {
int cnt=0;
queue<int>q;
for(int i=1; i<=n; i++) if(in[i]==0) q.push(i);
while(!q.empty()) {
int u=q.front();
q.pop();
cnt++;
for(auto v:e[u]) {
in[v]--;
if(in[v]==0) q.push(v);
}
}
if(cnt!=n) return true;
else return false;
}
int main() {
IOS;
#ifdef xiaofan
freopen("1.in","r",stdin);
freopen("1.out","w",stdout);
#endif
cin>>n>>m;
for(int i=1; i<=m; i++) {
cin>>a[i]>>b[i];
e[a[i]].push_back(b[i]);
in[b[i]]++;
}
if(tp()) {
cout<<2<<endl;
for(int i=1; i<=m; i++) cout<<(a[i]>b[i]?1:2)<<" ";
} else {
cout<<1<<endl;
for(int i=1; i<=m; i++) cout<<1<<" ";
}
return 0;
}