UVA 1108 Mining Your Own Business(双联通分量)
2016-11-13 10:20
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这题不能在割点上放救生装置很容易想到,关键就是如果整个图都是一个双联通的话,那么就需要放两个,因为放一个的话有可能放的那个会坍塌。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stack> #include<vector> using namespace std; const int maxn=50000+5; struct edge{int u,v;}; int pre[maxn],iscut[maxn],bccno[maxn],dfs_clock,bcc_cnt; vector<int> G[maxn],bcc[maxn]; stack<edge> s; int n,m; int dfs(int u,int fa) { int lowu=pre[u]=++dfs_clock; int child=0; for(int i=0;i<G[u].size();i++){ int v=G[u][i]; edge e=(edge){u,v}; if(!pre[v]){ s.push(e); child++; int lowv=dfs(v,u); lowu=min(lowu,lowv); if(lowv>=pre[u]){ iscut[u]=1; bcc_cnt++;bcc[bcc_cnt].clear(); for(;;){ edge x=s.top();s.pop(); if(bccno[x.u]!=bcc_cnt) {bcc[bcc_cnt].push_back(x.u);bccno[x.u]=bcc_cnt;} if(bccno[x.v]!=bcc_cnt) {bcc[bcc_cnt].push_back(x.v);bccno[x.v]=bcc_cnt;} if(x.u==u&&x.v==v) break; } } } else if(pre[v]<pre[u]&&v!=fa){ s.push(e); lowu=min(lowu,pre[v]); } } if(fa<0&&child==1) iscut[u]=0; return lowu; } void findbcc(){ memset(pre,0,sizeof(pre)); memset(iscut,0,sizeof(iscut)); memset(bccno,0,sizeof(bccno)); dfs_clock=0;bcc_cnt=0; for(int i=0;i<n;i++) if(!pre[i]) dfs(i,-1); } int main() { int kase=0; while(scanf("%d",&m)&&m) { for(int i=0;i<maxn;i++) G[i].clear(); n=-1; for(int i=0;i<m;i++){ int u,v; scanf("%d%d",&u,&v); n=max(n,max(u,v));u--;v--; G[u].push_back(v);G[v].push_back(u); } findbcc(); long long ans1=0,ans2=1; for(int i=1;i<=bcc_cnt;i++) { int cut_cnt=0; for(int j=0;j<bcc[i].size();j++) if(iscut[bcc[i][j]]) cut_cnt++; if(cut_cnt==1) {ans1++;ans2*=(bcc[i].size()-1);} } if(bcc_cnt==1) {ans1=2;ans2=bcc[1].size()*(bcc[1].size()-1)/2;} printf("Case %d: %lld %lld\n",++kase,ans1,ans2); } return 0; }
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