http://acm.hdu.edu.cn/showproblem.php?pid=4612
将原图进行缩点 变成一个树
树上每条边都是一个桥 然后加一条边要加在树的直径两端才最优
代码:
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<set>
#include<map>
#include<stack>
#include<vector>
#include<algorithm>
#include<queue>
#include<bitset>
#include<deque>
#include<numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef pair<int,int> pp;
const double eps=1e-9;
const int INF=0x3f3f3f3f;
const ll MOD=1000000007;
const int N=210000;
const int M=2100000;
int head1[N],I1;
int head2[N],I2;
struct node
{
int j,next;
}edge1[M],edge2[M];
bool ev[M];
int dist[N];
int low[N],dfn[N],f[N],deep;
bool in[N],visited[N];
int num[N],sel[N];
stack<int>st;
queue<int>qt;
vector<int>vt[N];
void add1(int i,int j)
{
edge1[I1].j=j;
edge1[I1].next=head1[i];
head1[i]=I1++;
}
void add2(int i,int j)
{
edge2[I2].j=j;
edge2[I2].next=head2[i];
head2[i]=I2++;
}
void tarjan(int x)//将环缩点
{
visited[x]=true;
in[x]=true;
st.push(x);
low[x]=dfn[x]=deep++;
for(int t=head1[x];t!=-1;t=edge1[t].next)
if(!ev[t])
{
ev[t]=true;
ev[t^1]=true;
int j=edge1[t].j;
if(visited[j]==false)
{
tarjan(j);
low[x]=min(low[x],low[j]);
}else if(in[j]==true)
{
low[x]=min(low[x],dfn[j]);
}
}
if(low[x]==dfn[x])
{
while(st.top()!=x)
{
int k=st.top(); st.pop();
in[k]=false;
f[k]=x;
vt[x].push_back(k);
}
int k=st.top(); st.pop();
in[k]=false;
f[k]=x;
}
}
void buildNewGraph(int k)
{
memset(head2,-1,sizeof(head2));I2=0;
for(int i=0;i<k;++i)
{
for(int t=head1[i];t!=-1;t=edge1[t].next)
{
int j=edge1[t].j;
if(f[i]!=f[j])
{
add2(f[j],f[i]);
++num[f[i]];
}
}
}
}
void init(int n,int m)
{
memset(head1,-1,sizeof(head1));
I1=0;
while(m--)
{
int l,r;
scanf("%d %d",&l,&r);
--l;--r;
add1(l,r);
add1(r,l);
}
}
int bfs(int x1,int n)
{
memset(dist,-1,sizeof(dist));
queue<int>qt;
dist[x1]=0;
qt.push(x1);
while(!qt.empty())
{
int x=qt.front();
qt.pop();
for(int t=head2[x];t!=-1;t=edge2[t].next)
{
int j=edge2[t].j;
if(dist[j]==-1)
{
dist[j]=dist[x]+1;
qt.push(j);
}
}
}
int k=0;
for(int i=0;i<n;++i)
if(dist[i]>dist[k])
k=i;
return k;
}
int solve(int n)
{
int s=0,sum=0;
for(int i=0;i<n;++i)
if(f[i]==i)
{s=i;++sum;}
int k=bfs(s,n);
k=bfs(k,n);
return (sum-1-dist[k]);
}
int main()
{
//freopen("data.in","r",stdin);
int n,m;
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0&&m==0)
break;
init(n,m);
while(!st.empty()) st.pop();
for(int i=0;i<n;++i)
{vt[i].clear();f[i]=i;}
memset(in,false,sizeof(in));
memset(visited,false,sizeof(visited));
memset(ev,false,sizeof(ev));
deep=0;
for(int i=0;i<n;++i)
if(!visited[i])
tarjan(i);
buildNewGraph(n);
printf("%d\n",solve(n));
}
return 0;
}
浙公网安备 33010602011771号