【连通图|边双连通+缩点】POJ-3177 Redundant Paths

”一个有桥的连通图，如何把它通过加边变成边双连通图？方法为首先求出所有的桥，然后删除这些桥边，剩下的每个连通块都是一个双连通子图。把每个双连通子图收缩为一个顶点，再把桥边加回来，最后的这个图一定是一棵树，边连通度为1。

统计出树中度为1的节点的个数，即为叶节点的个数，记为leaf。则至少在树上添加(leaf+1)/2条边，就能使树达到边双连通，所以至少添加的边数就是(leaf+1)/2。具体方法为，首先把两个最近公共祖先最远的两个叶节点之间连接一条边，这样可以把这两个点到祖先的路径上所有点收缩到一起，因为一个形成的环一定是双连通的。然后再找两个最近公共祖先最远的两个叶节点，这样一对一对找完，恰好是(leaf+1)/2次，把所有点收缩到了一起。“ --BYVoid

/*
 * ID: j.sure.1
 * PROG:
 * LANG: C++
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <climits>
#include <iostream>
#define Mem(f, x) memset(f, x, sizeof(f))
#define PB push_back
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
/****************************************/
const int N = 5e3+5, M = 2e4+5;
struct Edge {
    int v, next, idx, tag;
    Edge(){}
    Edge(int _v, int _next, int _idx, int _tag = 0):
        v(_v), next(_next), idx(_idx), tag(_tag){}
}e[M];
int n, m, deep, tot, bcc_cnt;
int dfn[N], head[N], bcc_id[N], line[N][2], deg[N];
bool isbri[M];

void __init__()
{
    bcc_cnt = tot = deep = 0;
    for(int i = 0; i < n; i++) {
        head[i] = -1;
        deg[i] = dfn[i] = bcc_id[i] = 0;
    }
    Mem(isbri, 0);
}

void add(int u, int v, int idx)
{
    int p;
    for(p = head[u]; ~p; p = e[p].next) {
        if(e[p].v == v) {
            e[p].tag++;
            return ;
        }
    }
    e[tot] = Edge(v, head[u], idx);
    head[u] = tot++;
}

int dfs(int u, int fa)
{
    int lowu = dfn[u] = ++deep;
    for(int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].v;
        if(!dfn[v]) {
            int lowv = dfs(v, u);
            lowu = min(lowu, lowv);
            if(lowv > dfn[u] && !e[i].tag) 
                isbri[e[i].idx] = 1;
        }
        else if(dfn[v] < dfn[u] && v != fa)
            lowu = min(lowu, dfn[v]);
    }
    return lowu;
}

void flood(int u) 
{
    bcc_id[u] = bcc_cnt;
    for(int i = head[u]; ~i; i = e[i].next) {
        if(isbri[e[i].idx]) continue ;
        int v = e[i].v;
        if(!bcc_id[v]) {
            flood(v);
        }
    }
}

int main()
{
#ifdef J_Sure
    freopen("000.in", "r", stdin);
    //freopen("999.out", "w", stdout);
#endif
    while(~scanf("%d%d", &n, &m)) {
        int ID = 0;
        __init__();
        for(int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            u--; v--;
            line[i][0] = u; line[i][1] = v;
            add(u, v, ID);
            add(v, u, ID++);
        }
        dfs(0, -1);
        for(int i = 0; i < n; i++) {
            if(!bcc_id[i]) {
                bcc_cnt++;
                flood(i);
            }
        }
        for(int i = 0; i < m; i++) {
            int u = bcc_id[line[i][0]], v = bcc_id[line[i][1]];
            if(u != v) {
                deg[u]++; deg[v]++;
            }
        }
        int ans = 0;
        for(int i = 1; i <= bcc_cnt; i++) {
            if(deg[i] == 1) ans++;
        }
        printf("%d\n", (ans+1)/2);
    }
    return 0;
}