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| #include<iostream>
#include<cstdio>
#include<vector>
#include<stack>
#include<cstring>
using namespace std;
long long read()
{
long long x=0,f=1; char c=getchar();
while(!isdigit(c)){if(c=='-') f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
const int N=100000+100;
struct road
{
int to,IsBack;
road (int A,int B)
{
to=A,IsBack=B;
}
};
vector <int> e[N];
vector <road> e2[N];
int belong[N],nd_tot,nd_to,low[N],dfn[N],InStack[N],cnt[N];
stack <int> st;
void Tarjan(int now)
{
low[now]=dfn[now]=++nd_to;
InStack[now]=true;
st.push(now);
for(int i=0;i<int(e[now].size());i++)
if(dfn[e[now][i]]==0)
{
Tarjan(e[now][i]);
low[now]=min(low[now],low[e[now][i]]);
}
else if(InStack[e[now][i]]==true)
low[now]=min(low[now],low[e[now][i]]);
if(low[now]==dfn[now])
{
nd_tot++;
while(st.empty()==false)
{
int temp=st.top();
st.pop();
belong[temp]=nd_tot;
InStack[temp]=false;
cnt[nd_tot]++;
if(temp==now)
break;
}
}
}
int n,m,S,f[N][2];
int dfs(int now,int back)
{
if(f[now][back]>=0) return f[now][back];
int t_ans=0;
bool OK=false;
for(int i=0;i<int(e2[now].size());i++)
if(e2[now][i].to!=S and back-e2[now][i].IsBack>=0)
t_ans=max(t_ans,dfs(e2[now][i].to,back-e2[now][i].IsBack));
else if(back>=e2[now][i].IsBack)
OK=true;
if(t_ans!=0 or OK==true)
return f[now][back]=t_ans+cnt[now];
else
return f[now][back]=0;
}
int main()
{
n=read(),m=read();
for(int i=1;i<=n;i++)
e[i].reserve(4),
e2[i].reserve(4);
for(int i=1;i<=m;i++)
{
int s=read(),t=read();
e[s].push_back(t);
}
for(int i=1;i<=n;i++)
if(dfn[i]==0)
Tarjan(i);
S=belong[1];
for(int i=1;i<=n;i++)
for(int j=0;j<int(e[i].size());j++)
if(belong[i]!=belong[e[i][j]])
{
e2[belong[i]].push_back(road(belong[e[i][j]],0));
e2[belong[e[i][j]]].push_back(road(belong[i],1));
}
memset(f,0x80,sizeof f);
int ans=0;
for(int i=0;i<int(e2[S].size());i++)
ans=max(ans,dfs(e2[S][i].to,1-e2[S][i].IsBack));
printf("%d",ans+cnt[S]);
return 0;
}
|
如果x能到达1所在的强连通分量或max出来的值不为0，说明当前状态可行，否则不可行。 然后用记忆化搜索表达出来就OK了
