AT5256 [ABC142F] Pure(有向图求最小环)
思路:
枚举每个点作为原点,求最小环长度。
关于求最小环,将spfa算法改为:将与原点所连的所有点放入队列。
输出方案的话,维护前驱数组记录是如何转移的。
推荐一篇很好的博客
代码:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
ll res = 1;
while(b)
{
if(b & 1)res = res * a % p;
a = a * a % p;
b >>= 1;
}
return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=1100;
vector<int>g[maxn];
int n,m,mp[maxn][maxn],dis[maxn],pre[maxn],res[maxn],st[maxn];
int spfa(int s){
memset(st,0,sizeof st);
queue<int>q;
memset(dis,0x3f,sizeof dis);
for(int i=0;i<g[s].size();i++){
int j=g[s][i];
pre[j]=s;
st[j]=1;
dis[j]=1;
q.push(j);
}
while(!q.empty()){
int t=q.front();q.pop();
st[t]=0;
for(int i=0;i<g[t].size();i++){
int j=g[t][i];
if(dis[j]>dis[t]+1){
dis[j]=dis[t]+1;
pre[j]=t;
if(!st[j]){
st[j]=1;
q.push(j);
}
}
}
}
return dis[s];
}
int main()
{
n=read,m=read;
rep(i,1,m){
int u=read,v=read;
g[u].push_back(v);
mp[u][v]=1;
}
int minn=inf,s=-1;
for(int i=1;i<=n;i++){
int tmp=spfa(i);
if(tmp<minn){
minn=tmp;
s=i;
for(int j=1;j<=n;j++) res[j]=pre[j];
}
}
if(minn==inf){
puts("-1");
}
else{
printf("%d\n",minn);
vector<int>path;
int now=s;
while(res[now]!=s){
path.push_back(now);
now=res[now];
}
path.push_back(now);
reverse(path.begin(),path.end());
for(int i=0;i<path.size();i++)
cout<<path[i]<<endl;
}
return 0;
}
/*
**/