The output file should contain a single line with space-delimited integers x and y, the IDs of servers to be connected by the new link..
1 ≤ n ≤ 10 000; 1≤ m ≤ 100 000; 1 ≤ xi, yi ≤ n; xi ≠ yi.
1 #pragma comment(linker, "/STACK:102400000,102400000")
2 #include<iostream>
3 #include<stdio.h>
4 #include<string.h>
5 const int VM = 200005;
6 const int EM = 1000005;
7
8 struct Edeg
9 {
10 int to,nxt,vis;
11 }edge[EM<<1],tree[EM<<1];
12
13 int head[VM],vis[VM],thead[VM];
14 int dfn[VM],low[VM],stack[VM],belong[VM];
15 int ep,bridge,son,maxn,src,n,cnt,scc,top;
16 int te[VM];
17 int max (int a,int b)
18 {
19 return a > b ? a : b;
20 }
21 int min(int a ,int b)
22 {
23 return a > b ? b : a;
24 }
25 void addedge (int cu,int cv)
26 {
27 edge[ep].to = cv;
28 edge[ep].vis = 0;
29 edge[ep].nxt = head[cu];
30 head[cu] = ep ++;
31 edge[ep].to = cu;
32 edge[ep].vis = 0;
33 edge[ep].nxt = head[cv];
34 head[cv] = ep ++;
35 }
36 void Buildtree(int cu,int cv)
37 {
38 tree[son].to = cv;
39 tree[son].nxt = thead[cu];
40 thead[cu] = son ++;
41 }
42 void Tarjan (int u)
43 {
44 int v;
45 vis[u] = 1;
46 dfn[u] = low[u] = ++cnt;
47 stack[top++] = u;
48 for (int i = head[u];i != -1;i = edge[i].nxt)
49 {
50 v = edge[i].to;
51 if (edge[i].vis) continue; //
52 edge[i].vis = edge[i^1].vis = 1; //正向边访问过了，反向边得标志，否则两点会成一块。
53 if (vis[v] == 1)
54 low[u] = min(low[u],dfn[v]);
55 if (!vis[v])
56 {
57 Tarjan (v);
58 low[u] = min(low[u],low[v]);
59 if (low[v] > dfn[u])
60 bridge ++;
61 }
62 }
63 if (dfn[u] == low[u])
64 {
65 ++scc;
66 do{
67 v = stack[--top];
68 vis[v] = 0;
69 belong[v] = scc;
70 te[scc]=v;
71 }while (u != v);
72 }
73 }
74 void BFS(int u)
75 {
76 int que[VM+100];
77 int front ,rear,i,v;
78 front = rear = 0;
79 memset (vis,0,sizeof(vis));
80 que[rear++] = u;
81 vis[u] = 1;
82 while (front != rear)
83 {
84 u = que[front ++];
85 front = front % (n+1);
86 for (i = thead[u];i != -1;i = tree[i].nxt)
87 {
88 v = tree[i].to;
89 if (vis[v]) continue;
90 vis[v] = 1;
91 que[rear++] = v;
92 rear = rear%(n+1);
93 }
94 }
95 src = que[--rear];//求出其中一个端点
96 }
97 int ans2=0;
98 void DFS (int u,int dep)
99 {
100 if(maxn<dep)
101 {
102 maxn = max(maxn,dep);
103 ans2 = u;
104 }
105 vis[u] = 1;
106 for (int i = thead[u]; i != -1; i = tree[i].nxt)
107 {
108 int v = tree[i].to;
109 if (!vis[v])
110 DFS (v,dep+1);
111 }
112 }
113 void solve()
114 {
115 int u,v;
116 memset (vis,0,sizeof(vis));
117 cnt = bridge = scc = top = 0;
118 Tarjan (1);
119 memset (thead,-1,sizeof(thead));
120 son = 0;
121 for (u = 1;u <= n;u ++) //重构图
122 for (int i = head[u];i != -1;i = edge[i].nxt)
123 {
124 v = edge[i].to;
125 if (belong[u]!=belong[v])
126 {
127 Buildtree (belong[u],belong[v]);
128 Buildtree (belong[v],belong[u]);
129 }
130 }
131 maxn = 0; //最长直径
132 BFS(1); //求树直径的一个端点
133 memset (vis,0,sizeof(vis));
134 DFS(src,0); //求树的最长直径
135 printf("%d %d\n",te[src],te[ans2]);
136 //printf ("%d\n",bridge-maxn);
137 }
138
139 int main ()
140 {
141 freopen("input.txt","r",stdin);
142 freopen("output.txt","w",stdout);
143 int m,u,v;
144 while (~scanf ("%d%d",&n,&m))
145 {
146 if (n == 0&&m == 0)
147 break;
148 memset (head,-1,sizeof(head));
149 ep = 0;
150 while (m --)
151 {
152 scanf ("%d%d",&u,&v);
153 addedge (u,v);
154 }
155 solve();
156 }
157 return 0;
158 }
