E. Minimal Diameter Forest
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a forest — an undirected graph with nn vertices such that each its connected component is a tree.
The diameter (aka "longest shortest path") of a connected undirected graph is the maximum number of edges in the shortest path between any pair of its vertices.
You task is to add some edges (possibly zero) to the graph so that it becomes a tree and the diameter of the tree is minimal possible.
If there are multiple correct answers, print any of them.
Input
The first line contains two integers nn and mm (1≤n≤10001≤n≤1000, 0≤m≤n−10≤m≤n−1) — the number of vertices of the graph and the number of edges, respectively.
Each of the next mm lines contains two integers vv and uu (1≤v,u≤n1≤v,u≤n, v≠uv≠u) — the descriptions of the edges.
It is guaranteed that the given graph is a forest.
Output
In the first line print the diameter of the resulting tree.
Each of the next (n−1)−m(n−1)−m lines should contain two integers vv and uu (1≤v,u≤n1≤v,u≤n, v≠uv≠u) — the descriptions of the added edges.
The resulting graph should be a tree and its diameter should be minimal possible.
For m=n−1m=n−1 no edges are added, thus the output consists of a single integer — diameter of the given tree.
If there are multiple correct answers, print any of them.
Examples
input
Copy
4 2 1 2 2 3
output
Copy
2 4 2
input
Copy
2 0
output
Copy
1 1 2
input
Copy
3 2 1 3 2 3
output
Copy
2
Note
In the first example adding edges (1, 4) or (3, 4) will lead to a total diameter of 3. Adding edge (2, 4), however, will make it 2.
Edge (1, 2) is the only option you have for the second example. The diameter is 1.
You can't add any edges in the third example. The diameter is already 2.
题意：给你一个含n(n<=1000)个节点、m(<n)条边的无向无环无重边图（森林），让你添加边将其连接成一棵树，使树的直径最小，输出最后树的最短直径，和你添加的边。
其实思路很好想，明显是求每颗树的直径的链的中点，然后用最长的那棵树的直径的中点连接其他树的中点。（想想是不是）
关键是怎么写。。。
理清思路，先求出每个连通块的点、中心，同时求出最长直径的树及其连通块编号，然后添加边构成大树，然后再求一遍直径。这个过程可以用若干个dfs来实现，注意细节。
代码：
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1010;
vector<int>vc[maxn],cv;
set<int>st;
set<int>::iterator it,itt;
int n,m,k,id;
int ans,tmp,cnt,ct,lid;
int ansid,ansl;
int vis[maxn],idx[maxn],du[maxn];
int in[maxn];
void dfs(int u,int fa)
{
vis[u]=1;
st.insert(u);
in[u]=1;
for(int i=0;i<vc[u].size();i++)
{
int v=vc[u][i];
if(v!=fa) dfs(v,u);
}
}
void findc()
{
vector<int>vv;
for(it=st.begin();it!=st.end();)
{
itt=it;
it++;
int u=*itt;
if(du[u]==1)
{
vv.push_back(u);
in[u]=0;
if(st.size()>1)st.erase(u);
}
}
//cout<<(*st.begin())<<endl;
while(st.size()>2)
{
//cout<<" I "<<(*it)<<endl;
vector<int>v1;
for(int i=0;i<vv.size();i++)
{
int u=vv[i];
for(int j=0;j<vc[u].size();j++)
if(in[vc[u][j]]){
if(--du[vc[u][j]]==1)
{
v1.push_back(vc[u][j]);
st.erase(vc[u][j]);
in[vc[u][j]]=0;
}
}
}
vv=v1;
}
//cout<<" * "<<(*st.begin())<<endl;
cv.push_back(*st.begin());
}
void dfs2(int u,int fa,int dep)
{
if(dep>ans)
{
ans=dep;
lid=u;
}
for(int i=0;i<vc[u].size();i++)
{
int v=vc[u][i];
if(v==fa) continue;
dfs2(v,u,dep+1);
}
}
int findzj(int u=1,int idd=1)
{
ans=-1;
dfs2(u,0,0);
ans=-1;
dfs2(lid,0,0);
if(idd&&ans>ansl)
{
ansl=ans;
ansid=id;
}
return ans;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
int u,v;
scanf("%d%d",&u,&v);
vc[u].push_back(v);
vc[v].push_back(u);
du[u]++;
du[v]++;
}
id=1;
ansl=-1;
for(int i=1;i<=n;i++)
if(!vis[i]) {
st.clear();
for(int j=1;j<=n;j++)
in[j]=0;
dfs(i,0);
findc();
findzj(i);
id++;
}
//cout<<ansid<<" "<<ansl<<" "<<cv.size()<<endl;
for(int i=0;i<cv.size();i++)
if(i==ansid-1) continue;
else {
int u=cv[i];
int v=cv[ansid-1];
vc[u].push_back(v);
vc[v].push_back(u);
}
printf("%d\n",findzj(1,0));
for(int i=0;i<cv.size();i++)
if(i==ansid-1) continue;
else {
int u=cv[i];
int v=cv[ansid-1];
printf("%d %d\n",u,v);
}
return 0;
}