hdu 4612 求边连通分量+求树的直径

http://acm.hdu.edu.cn/showproblem.php?pid=4612
题意：
加一条边，使得桥的数量最少？
思路：
求边连通分量，缩点，然后变成一个树，树边都是桥，求树的直径。
树的直径就是最多减少的桥。
桥的数量最少就是原图中的桥的数量 - 树的直径。

要注意重边。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef pair<int,int> pii;
const int M_node = 200009, M_edge = 2000009;
int low[M_node],dfn[M_node],belong[M_node],head[M_node];
int dfs_clock,block,bridge,tot;
int n,m,maxlen,s;
bool instack[M_node];
struct Node
{
    int u,v;
}node[M_edge];
struct edge
{
    int to,next;
    bool cut;
    bool mul;
}edge[M_edge];
vector<int> G[M_node];
stack<int> st;
void init()
{
    memset(head,-1,sizeof(head));
    dfs_clock = block = bridge = tot = 0;
}
void add_edge(int u,int v,bool mul)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].cut = false;
    edge[tot].mul = mul;
    head[u] = tot++;
}
void tarjan(int u,int fa,bool mul)
{
    low[u] = dfn[u] = ++dfs_clock;
    st.push(u);
    instack[u] = true;
    for(int i = head[u]; i != -1;i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == fa && !mul) continue; //如果不是重边才跳过
        if(!dfn[v])
        {
            tarjan(v,u,edge[i].mul);
            low[u] = min(low[v],low[u]);
            if(low[v] > dfn[u])
            {
                bridge++;
                edge[i].cut = true;
                edge[i^1].cut = true;
            }
        }
        else if(instack[u] && low[u] > dfn[v]) low[u] = dfn[v];
    }
    if(low[u] == dfn[u])
    {
        block++;
        for(;;)
        {
            int v = st.top();
            st.pop();
            instack[v] = false;
            belong[v] = block;
            if(v == u) break;
        }
    }
}
void dfs(int u,int pre,int len)
{
    if(maxlen < len)
    {
        s = u;
        maxlen = len;
    }
    for(int i = 0; i < G[u].size();i++)
    {
        int v = G[u][i];
        if(v == pre) continue;
        dfs(v,u,len+1);
    }
}
void solve()
{
    while(!st.empty()) st.pop();
    memset(dfn,0,sizeof(dfn));
    memset(instack,false,sizeof(instack));
    tarjan(1,-1,false);
    for(int i = 1;i <= block;i++) G[i].clear();
    for(int i = 1;i <= n;i++)
    {
        for(int j = head[i];j != -1;j = edge[j].next)
        {
            if(edge[j].cut)
            {
                //printf("debug -- i = %d , belong[i] = %d , edge[j].to = %d , belong[edge[j].to] = %d\n",i,belong[i],edge[j].to,belong[edge[j].to]);
                G[belong[i]].push_back(belong[edge[j].to]);
            }
        }
    }
    maxlen = -1;
    dfs(1,-1,0);
    maxlen = -1;
    dfs(s,-1,0);
    printf("%d\n",bridge - maxlen);
}
bool cmp(Node a,Node b)
{
    if(a.u == b.u) return a.v < b.v;
    return a.u < b.u;
}
int main()
{
    while(scanf("%d %d",&n,&m) == 2)
    {
        init();
        if(n == 0 && m == 0) break;
        for(int i = 0;i < m;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            if(a == b) continue;
            if(a > b) swap(a,b);
            node[i].u = a;
            node[i].v = b;
        }
        sort(node,node+m,cmp);
        for(int i = 0;i < m;i++)
        {
            if(i == 0 || (node[i].u != node[i-1].u || node[i].v != node[i-1].v))
            {
                if(i < m - 1 && node[i].u == node[i+1].u && node[i].v == node[i+1].v)
                {
                    add_edge(node[i].u,node[i].v,true);
                    add_edge(node[i].v,node[i].u,true);
                }
                else
                {
                    add_edge(node[i].u,node[i].v,false);
                    add_edge(node[i].v,node[i].u,false);
                }
            }
        }
        solve();
    }
    return 0;
}