Redundant Paths 分離的路徑

題目描述

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.

為了從F(1≤F≤5000)個草場中的一個走到另一個，貝茜和她的同伴們有時不得不路過一些她們討厭的可怕的樹．奶牛們已經厭倦了被迫走某一條路，所以她們想建一些新路，使每一對草場之間都會至少有兩條相互分離的路徑，這樣她們就有多一些選擇．
每對草場之間已經有至少一條路徑．給出所有R(F-1≤R≤10000)條雙向路的描述，每條路連線了兩個不同的草場，請計算最少的新建道路的數量, 路徑由若干道路首尾相連而成．兩條路徑相互分離，是指兩條路徑沒有一條重合的道路．但是，兩條分離的路徑上可以有一些相同的草場． 對於同一對草場之間，可能已經有兩條不同的道路，你也可以在它們之間再建一條道路，作為另一條不同的道路．

輸入格式

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行輸入F和R，接下來R行，每行輸入兩個整數，表示兩個草場，它們之間有一條道路．

輸出格式

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數．

樣例

樣例輸入

7 7
1 2
2 3
3 4
2 5
4 5
5 6
5 7

樣例輸出

2

資料範圍與提示

思路分析

裸裸的邊雙聯通分量（但是我不會寫），借這個題詳解一下

關於邊雙聯通分量：

• 定義：若一個無向圖中的去掉任意一條邊都不會改變此圖的連通性，即不存在橋，則稱作邊雙連通圖。一個無向圖中的每一個極大邊雙連通子圖稱作此無向圖的邊雙連通分量。

• 橋：連線兩個邊雙連通分量的邊即是橋。

求法:

• 基本思路： 一個非常神奇的思路就是無向圖縮點，因為每一個邊雙聯通分量都可以看做是一個環，那麼每一個換上的點都有兩個方向可以走，自然就不會存在橋了

• 關鍵點在於，既然是一個無向圖，那麼一條邊要變為兩條有向邊加入，但本質上確實一條邊，若不處理會容易重複，那麼怎麼維護他們的關係？

• 大佬發威（當然，不是我）：

  解決的方法就是，當同一條無向邊的兩條有向邊的其中一條走過時，把另一條同時賦值為走過，這就要用到一個神奇的公式，^1。 舉例來說，0^1=1,11=0; 2^1=3,31=2; 4^1=3,31=4......這樣正好每次加邊時兩條相鄰的邊就可以同時處理了，妙啊

程式碼

#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){
    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]){  這裡需要注意一下，是邊，而不是點
            vis[i] = vis[i^1] = 1;
            if(!dfn[v]){  然後再處理邊上的點
                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]]]++; 不在同一個邊雙裡就需要搭橋
        }
    }
    for(int i = 1;i <= col;i++){
        if(du[i]==1)ans++;
    }
    printf("%d\n",ans+1>>1);  兩個邊雙搭一個橋就夠了
}

髮量減1%