給一個有向圖，而後選一條路徑起點終點都爲1的路徑出來，有一次機會能夠沿某條邊逆方向走，問最多有多少個點能夠被通過？（一個點在路徑中不管出現多少正整數次對答案的貢獻均爲1）git
=>有向圖咱們先考慮縮點。而後觀察縮點後的圖能夠發現新的路徑中一定只有一條邊是反向的才符合條件。那麼咱們能夠聯想到某道最短路的題將邊反向存一遍後分別從s和t跑一跑。那麼這裏bfs跑一跑就好了。而後有一個坑點：這種重建圖的注意es和edges否則es會在中途就被修改掉了。。。 app
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define clr(x,c) memset(x,c,sizeof(x))
#define qwq(x) for(edge *o=head[x];o;o=o->next)
#define qaq(x) for(edge *o=hd[x];o;o=o->next)
#define TAT(x) for(edge *o=eo[x];o;o=o->next)
int read(){
int x=0;char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=x*10+c-'0',c=getchar();
return x;
}
const int nmax=1e5+5;
const int inf=0x7f7f7f7f;
struct edge{
int to;edge *next;
};edge es[nmax],edges[nmax<<1],*pt=es,*head[nmax],*hd[nmax],*eo[nmax];
void add(int u,int v){
pt->to=v;pt->next=head[u];head[u]=pt++;
}
void adde(int u,int v){
pt->to=v;pt->next=hd[u];hd[u]=pt++;
pt->to=u;pt->next=eo[v];eo[v]=pt++;
}
int pre[nmax],dfs_clock=0,scc_cnt=0,sccno[nmax],sm[nmax];
stack<int>s;
int dfs(int x){
int lowu=pre[x]=++dfs_clock;s.push(x);
qwq(x){
if(!pre[o->to]) lowu=min(lowu,dfs(o->to));
else if(!sccno[o->to]) lowu=min(lowu,pre[o->to]);
}
if(lowu==pre[x]){
++scc_cnt;int tc=0;
while(1){
int tx=s.top();s.pop();
sccno[tx]=scc_cnt;++tc;
if(x==tx) break;
}
sm[scc_cnt]=tc;
}
return lowu;
}
queue<int>q;bool vis[nmax];int f[nmax],g[nmax];
int test_cnt=0;
void bfs1(int x){
q.push(x);f[x]=sm[x];int tx;clr(vis,0);
while(!q.empty()){
tx=q.front();q.pop();vis[tx]=0;
qaq(tx) if(f[o->to]<f[tx]+sm[o->to]){
f[o->to]=f[tx]+sm[o->to];
if(!vis[o->to]) q.push(o->to),vis[o->to]=1;
}
}
}
void bfs2(int x){
q.push(x);g[x]=sm[x];int tx;clr(vis,0);
while(!q.empty()){
tx=q.front();q.pop();vis[tx]=0;
TAT(tx) if(g[o->to]<g[tx]+sm[o->to]){
g[o->to]=g[tx]+sm[o->to];
if(!vis[o->to]) q.push(o->to),vis[o->to]=1;
}
}
}
int main(){
int n=read(),m=read(),u,v;
rep(i,1,m) u=read(),v=read(),add(u,v);
rep(i,1,n) if(!pre[i]) dfs(i);
//rep(i,1,n) printf("%d ",sccno[i]);printf("\n");
pt=edges;
rep(i,1,n) qwq(i) if(sccno[i]!=sccno[o->to]) adde(sccno[i],sccno[o->to]);
bfs1(sccno[1]);bfs2(sccno[1]);
//rep(i,1,scc_cnt) printf("%d %d\n",f[i],g[i]);
int ans=sm[sccno[1]];
rep(i,1,scc_cnt) qaq(i) {
if(g[i]&&f[o->to]) ans=max(ans,g[i]+f[o->to]-sm[sccno[1]]);
}
printf("%d\n",ans);return 0;
}
In an effort to better manage the grazing patterns of his cows, Farmer John has installed one-way cow paths all over his farm. The farm consists of N fields, conveniently numbered 1..N, with each one-way cow path connecting a pair of fields. For example, if a path connects from field X to field Y, then cows are allowed to travel from X to Y but not from Y to X. Bessie the cow, as we all know, enjoys eating grass from as many fields as possible. She always starts in field 1 at the beginning of the day and visits a sequence of fields, returning to field 1 at the end of the day. She tries to maximize the number of distinct fields along her route, since she gets to eat the grass in each one (if she visits a field multiple times, she only eats the grass there once). As one might imagine, Bessie is not particularly happy about the one-way restriction on FJ's paths, since this will likely reduce the number of distinct fields she can possibly visit along her daily route. She wonders how much grass she will be able to eat if she breaks the rules and follows up to one path in the wrong direction. Please compute the maximum number of distinct fields she can visit along a route starting and ending at field 1, where she can follow up to one path along the route in the wrong direction. Bessie can only travel backwards at most once in her journey. In particular, she cannot even take the same path backwards twice.
給一個有向圖，而後選一條路徑起點終點都爲1的路徑出來，有一次機會能夠沿某條邊逆方向走，問最多有多少個點能夠被通過？（一個點在路徑中不管出現多少正整數次對答案的貢獻均爲1）this
The first line of input contains N and M, giving the number of fields and the number of one-way paths (1 <= N, M <= 100,000). The following M lines each describe a one-way cow path. Each line contains two distinct field numbers X and Y, corresponding to a cow path from X to Y. The same cow path will never appear more than once.
A single line indicating the maximum number of distinct fields Bessie
can visit along a route starting and ending at field 1, given that she can
follow at most one path along this route in the wrong direction.