POJ-3352 Redundant Paths

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.

Input

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.

Output

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

Sample Input

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

Sample Output

2

Hint

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.

 

题目所求即为把无向图中缩点为一棵树后，再加边使之成为一个边双连通块。所加边数即为(叶子节点数+1)/2，加边方法为每次取两个LCA最远的叶节点，在他们两个中间连一条边，重复取直到加完。

注意：这道题两点之间会有多条边，不能简单的判断 To ！= father ，必须判断是否为同一条边。因为边为无向边，所以不知道 i+1 和 i-1 那一条和 i 是同一条边。这里介绍一种妙不可言的方法，把每条无向边的编号赋值为这条边的权，所以只需判断两条边权是否相同即可。

 #include <map>
 #include <set>
 #include <cmath>
 #include <ctime>
 #include <queue>
 #include <stack>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 #define ll long long
 #define file(a) freopen(a".in","r",stdin); freopen(a".out","w",stdout);

 inline int gi()
 {
     bool b=; int r=; char c=getchar();
     while(c<'' || c>'') { if(c=='-') b=!b; c=getchar(); }
     while(c>='' && c<='') { r=r*+c-''; c=getchar(); }
     if(b) return -r; return r;
 }

 const int inf = 1e9+, N = , M = ;
 int n,m,num,Deep,f[N],dfn[N],low[N],cd[N];
 bool b[N];
 stack <int> s;
 struct data
 {
     int nx,to,ds;
 }da[M];

 inline void add (int fr,int to,int ds)
 {
     da[++num].to=to, da[num].nx=f[fr], da[num].ds=ds, f[fr]=num;
 }

 inline void tarjan (int o,int fa)
 {
     int i,to;
     dfn[o]=low[o]=++Deep; b[o]=;
     for (i=f[o]; i; i=da[i].nx)
         {
             to=da[i].to;
             if (da[i].ds == da[fa].ds) continue;
             if (!dfn[to]) tarjan (to,i), low[o]=min(low[o],low[to]);
             else if (b[to]) low[o]=min(low[o],dfn[to]);
         }
     b[o]=;
 }

 int main()
 {
 //    file("POJ-3177");
     n=gi(), m=gi();
     int i,j,x,y;
     for (i=; i<=m; i++)
         {
             x=gi(), y=gi();
             add (x,y,i), add (y,x,i);
         }
     for (i=; i<=n; i++) if (!dfn[i]) tarjan (i,);
     for (i=; i<=n; i++)
         for (j=f[i]; j; j=da[j].nx)
             {
                 x=low[da[j].to], y=low[i];
                 if (x != y) cd[y]++;
             }
     x=;
     for (i=; i<=n; i++) if (cd[i] == ) x++;
     printf ("%d\n",(x+)/);
     return ;
 }

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