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，每條路連接了兩個不同的草場，請計算最少的新建道路的數量, 路徑由若幹道路首尾相連而成．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．

輸入輸出格式

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.

輸出格式：

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

輸入輸出樣例

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

2

說明

Explanation of the sample:

One visualization of the paths is:

1 2 3

+---+---+

   |   |
   |   |

6 +---+---+ 4

/ 5 / / 7 +Building new paths from 1 to 6 and from 4 to 7 satisfies the conditions.

1 2 3

+---+---+

6 +---+---+ 4

/ 5 : / :

/ :

7 + - - - - Check some of the routes:

1 – 2: 1 –> 2 and 1 –> 6 –> 5 –> 2

1 – 4: 1 –> 2 –> 3 –> 4 and 1 –> 6 –> 5 –> 4

3 – 7: 3 –> 4 –> 7 and 3 –> 2 –> 5 –> 7

Every pair of fields is, in fact, connected by two routes.

It‘s possible that adding some other path will also solve the problem (like one from 6 to 7). Adding two paths, however, is the minimum.

首先，對於雙聯通分量裏的點我們是不用考慮的。

因此我們先用tarjan把雙聯通分量縮出來

這樣的話必定會形成一棵樹

接下來考慮貪心，對於任意三個不在同一雙聯通分量裏且入度唯一的點

我們在他們中間連一條邊，這樣一定是最優的

因此我們只需要統計入度為1的點就行了

最後的答案為$(ans+1)/2$

// luogu-judger-enable-o2
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#define getchar() (S == T && (T = (S = BB) + fread(BB, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
char BB[1 << 15], *S = BB, *T = BB;
using namespace std;
const int MAXN=1e5+10;
inline int read()
{
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
struct node
{
    int u,v,w,nxt;
}edge[MAXN];
int head[MAXN],num=1;
inline void AddEdge(int x,int y)
{
    edge[num].u=x;
    edge[num].v=y;
    edge[num].nxt=head[x];
    head[x]=num++;
}
int dfn[MAXN],low[MAXN],vis[MAXN],tot=0,colornum=0,color[MAXN],inder[MAXN];
int fuck[5001][5001];
stack<int>s;
void tarjan(int now,int fa)
{
    dfn[now]=low[now]=++tot;
    s.push(now);
    vis[now]=1;
    for(int i=head[now];i!=-1;i=edge[i].nxt)
    {
        if(!dfn[edge[i].v]&&edge[i].v!=fa) 
            tarjan(edge[i].v,now),low[now]=min(low[now],low[edge[i].v]);
        if(vis[edge[i].v]&&edge[i].v!=fa) low[now]=min(low[now],dfn[edge[i].v]);
    }
    if(dfn[now]==low[now])
    {
        int h=0;
        colornum++;
        do
        {
            h=s.top();
            color[h]=colornum;
            s.pop();
        }while(h!=now);
    }
}
int main()
{
    #ifdef WIN32
    freopen("a.in","r",stdin);
    #else
    #endif
    memset(head,-1,sizeof(head));
    int N=read(),M=read();
    for(int i=1;i<=M;i++)
    {
        int x=read(),y=read();
        AddEdge(x,y);
        AddEdge(y,x);
    }
    tarjan(1,0);
    for(int i=1;i<=num-1;i++) 
        if(color[edge[i].u]!=color[edge[i].v]&&fuck[edge[i].u][edge[i].v]==0)
            fuck[edge[i].u][edge[i].v]=1,
            inder[color[edge[i].u]]++,
            inder[color[edge[i].v]]++;
    int ans=0;
    for(int i=1;i<=colornum;i++)
        if(inder[i]==2)//雙向邊 
            ans++;
    printf("%d",(ans+1)>>1);    
    return 0;
}

洛谷P2860 [USACO06JAN]冗余路徑Redundant Paths(tarjan求雙聯通分量)