Redundant Paths 分离的路径

题目描述

In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two separate routes between any pair of fields. They currently have at least one route between each pair of fields and want to have at least two. Of course, they can only travel on Official Paths when they move from one field to another. Given a description of the current set of R (F-1 <= R <= 10,000) paths that each connect exactly two different fields, determine the minimum number of new paths (each of which connects exactly two fields) that must be built so that there are at least two separate routes between any pair of fields. Routes are considered separate if they use none of the same paths, even if they visit the same intermediate field along the way. There might already be more than one paths between the same pair of fields, and you may also build a new path that connects the same fields as some other path.

为了从F(1≤F≤5000)个草场中的一个走到另一个，贝茜和她的同伴们有时不得不路过一些她们讨厌的可怕的树．奶牛们已经厌倦了被迫走某一条路，所以她们想建一些新路，使每一对草场之间都会至少有两条相互分离的路径，这样她们就有多一些选择．

每对草场之间已经有至少一条路径．给出所有R(F-1≤R≤10000)条双向路的描述，每条路连接了两个不同的草场，请计算最少的新建道路的数量, 路径由若干道路首尾相连而成．两条路径相互分离，是指两条路径没有一条重合的道路．但是，两条分离的路径上可以有一些相同的草场．对于同一对草场之间，可能已经有两条不同的道路，你也可以在它们之间再建一条道路，作为另一条不同的道路．

输入格式

Line 1: Two space-separated integers: F and R

Lines 2..R+1: Each line contains two space-separated integers which are the fields at the endpoints of some path.

第1行输入F和R，接下来R行，每行输入两个整数，表示两个草场，它们之间有一条道路．

输出格式

Line 1: A single integer that is the number of new paths that must be built.

最少的需要新建的道路数．

样例

样例输入

样例输出

数据范围与提示

思路分析

裸裸的边双联通分量（但是我不会写），借这个题详解一下

关于边双联通分量：

定义：若一个无向图中的去掉任意一条边都不会改变此图的连通性，即不存在桥，则称作边双连通图。一个无向图中的每一个极大边双连通子图称作此无向图的边双连通分量。
桥：连接两个边双连通分量的边即是桥。

求法:

基本思路：一个非常神奇的思路就是无向图缩点，因为每一个边双联通分量都可以看做是一个环，那么每一个换上的点都有两个方向可以走，自然就不会存在桥了
关键点在于，既然是一个无向图，那么一条边要变为两条有向边加入，但本质上确实一条边，若不处理会容易重复，那么怎么维护他们的关系？
大佬发威（当然，不是我）：

解决的方法就是，当同一条无向边的两条有向边的其中一条走过时，把另一条同时赋值为走过，这就要用到一个神奇的公式，^1。举例来说，0 ^ 1=1,1 ^ 1=0; 2 ^ 1=3,3 ^ 1=2; 4 ^ 1=3,3 ^ 1=4......这样正好每次加边时两条相邻的边就可以同时处理了，妙啊

代码

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

const int N = 5e3+10,M = 1e4+10;

using namespace std;

inline int read(){

   int s=0,w=1;

   char ch=getchar();

   while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}

   while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();

   return s*w;

}

int n,m,vis[M<<1],du[N],ans;

int cnt=1,head[N],u[M],v[M];

int now,top,col,dfn[N],low[N],sta[N],color[N];

struct edge{

    int next,to;

}e[M<<1];

void add(int u,int v){  //注意我们cnt的初始值为1，这样保证边是从2,3这两个有相关关系的计数开始的

    e[++cnt].to = v;

    e[cnt].next = head[u];

    head[u] = cnt;

    e[++cnt].to = u;

    e[cnt].next = head[v];

    head[v] = cnt;

}

#define v e[i].to

void tarjan(int u){

    dfn[u]=low[u]=++now;

    sta[++top] = u;

    for(int i = head[u];i;i = e[i].next){

        if(!vis[i]){  这里需要注意一下，是边，而不是点

            vis[i] = vis[i^1] = 1;

            if(!dfn[v]){  然后再处理边上的点

                tarjan(v);

                low[u] = min(low[u],low[v]);

            }

            else low[u] = min(low[u],dfn[v]);

        }

    }

    if(low[u]==dfn[u]){

        color[u] = ++col;

        while(sta[top]!=u){

            color[sta[top--]] = col;

        }

        top--;

    }

}

#undef v

int main(){

    n =read();m = read();

    for(int i = 1;i <= m;i++){

        u[i] = read();v[i] = read();

        add(u[i],v[i]);

    }

    for(int i = 1;i <= n;i++){

        if(!dfn[i])tarjan(i);

    }

    for(int i = 1;i <= m;i++){

        if(color[u[i]] != color[v[i]]){

            du[color[u[i]]]++,du[color[v[i]]]++; 不在同一个边双里就需要搭桥

        }

    }

    for(int i = 1;i <= col;i++){

        if(du[i]==1)ans++;

    }

    printf("%d\n",ans+1>>1);  两个边双搭一个桥就够了

}

发量减1%