HDU 3844 Mining Your Own Business

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Mining Your Own BusinessTime Limit: 1000msMemory Limit: 32768KBThis problem will be judged on HDU. Original ID: 3844
64-bit integer IO format: %I64d      Java class name: Main John Digger is the owner of a large illudium phosdex mine. The mine is made up of a series of tunnels that meet at various large junctions. Unlike some owners, Digger actually cares about the welfare of his workers and has a concern about the layout of the mine. Specifically, he worries that there may a junction which, in case of collapse, will cut off workers in one section of the mine from other workers (illudium phosdex, as you know, is highly unstable). To counter this, he wants to install special escape shafts from the junctions to the surface. He could install one escape shaft at each junction, but Digger doesn’t care about his workers that much. Instead, he wants to install the minimum number of escape shafts so that if any of the junctions collapses, all the workers who survive the junction collapse will have a path to the surface.

Write a program to calculate the minimum number of escape shafts and the total number of ways in which this minimum number of escape shafts can be installed. InputThe input consists of several test cases. The first line of each case contains a positive integer N (N <= 5×10^4) indicating the number of mine tunnels. Following this are N lines each containing two distinct integers s and t, where s and t are junction numbers. Junctions are numbered consecutively starting at 1. Each pair of junctions is joined by at most a single tunnel. Each set of mine tunnels forms one connected unit (that is, you can get from any one junction to any other).

The last test case is followed by a line containing a single zero. OutputFor each test case, display its case number followed by the minimum number of escape shafts needed for the system of mine tunnels and the total number of ways these escape shafts can be installed. You may assume that the result fits in a signed 64-bit integer.

Follow the format of the sample output. Sample Input

91 34 13 51 22 61 56 31 63 261 21 32 42 53 63 70

Sample Output

Case 1: 2 4Case 2: 4 1

Source2011WorldFinal 解題：點雙連通分量，當然我不是直接求點雙連通分量的，我是求割點。先找出割點，然後從非割點開始dfs，在dfs的過程中，不能經過割點。 當一個點雙連通分量中有且僅有一個割點的時候才選擇一個位置進行安裝，安裝的可選的位置不包括割點。 當整個圖不存在割點的時候，需要兩個位置，因為如果一個，恰好放梯子的位置掛了，全掛了。所以隨便選兩個位置就好了。 此題目需要擴棧。。。。WF的題目滿是坑 

 1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <algorithm> 5 #pragma comment(linker, "/STACK:102400000,102400000") 6 using namespace std; 7 typedef long long LL; 8 const int maxn = 50010; 9 struct arc {10     int to,next;11     arc(int x = 0,int y = -1) {12         to = x;13         next = y;14     }15 } e[1000010];16 int head[maxn],dfn[maxn],low[maxn],idx,tot,cn;17 bool cut[maxn];18 void add(int u,int v) {19     e[tot] = arc(v,head[u]);20     head[u] = tot++;21     e[tot] = arc(u,head[v]);22     head[v] = tot++;23 }24 void tarjan(int u,int fa) {25     dfn[u] = low[u] = ++idx;26     int son = 0;27     for(int i = head[u]; ~i; i = e[i].next) {28         if(e[i].to == fa) continue;29         if(!dfn[e[i].to]) {30             tarjan(e[i].to,u);31             son++;32             low[u] = min(low[u],low[e[i].to]);33             if(fa != -1 && low[e[i].to] >= dfn[u] || fa == -1 && son > 1) {34                 cut[u] = true;35                 cn++;36             }37         } else low[u] = min(low[u],dfn[e[i].to]);38     }39 }40 bool vis[maxn];41 int cnt,n,m,cao;42 bool fk[maxn];43 void dfs(int u,int fa) {44     vis[u] = true;45     cnt++;46     for(int i = head[u]; ~i; i = e[i].next) {47         if(e[i].to == fa) continue;48         if(cut[e[i].to]) {49             if(!fk[e[i].to]) {50                 cao++;51                 fk[e[i].to] = true;52             }53             continue;54         }55         if(vis[e[i].to]) continue;56         dfs(e[i].to,fa);57     }58 }59 int main() {60     int u,v,cs = 1;61     while(scanf("%d",&m),m) {62         memset(head,-1,sizeof head);63         for(int i = tot = n = idx = cn = 0; i < m; ++i) {64             scanf("%d%d",&u,&v);65             add(u,v);66             n = max(n,max(u,v));67         }68         memset(dfn,0,sizeof dfn);69         memset(cut,false,sizeof cut);70         for(int i = 1; i <= n; ++i)71             if(!dfn[i]) tarjan(i,-1);72         LL ret = 1;73         if(!cn) printf("Case %d: %d %I64d\n",cs++,2,(LL)n*(n-1)/2);74         else {75             memset(vis,false,sizeof vis);76             int ri = 0;77             for(int i = 1; i <= n; ++i) {78                 if(cut[i] || vis[i]) continue;79                 cnt = cao = 0;80                 memset(fk,false,sizeof fk);81                 dfs(i,-1);82                 if(cao < 2 && cnt) {ri++; ret *= cnt;}83             }84             printf("Case %d: %d %I64d\n",cs++,ri,ret);85         }86     }87     return 0;88 }

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HDU 3844 Mining Your Own Business