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;
}

 

 

posted on 2013-07-25 21:27  夜->  阅读(246)  评论(0编辑  收藏  举报