hdu-4612(无向图缩点+树的直径)
题意:给你n个点和m条边的无向图,问你如果多加一条边的话,那么这个图最少的桥是什么
解题思路:无向图缩点和树的直径,用并查集缩点;
#include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<cstring> #include<queue> using namespace std; const int maxm=2005000; const int maxn=200500; struct Edge { int next;int id;int to; }edge[maxm]; int head[maxn],dist[maxn],f[maxn],low[maxn],dfn[maxn],visit[maxn]; int n,m,cnt,ans,step; void init() { memset(dist,0,sizeof(dist)); memset(visit,0,sizeof(visit)); memset(head,-1,sizeof(head)); memset(low,0,sizeof(low)); memset(dfn,0,sizeof(dfn)); for(int i=1;i<=n;i++) f[i]=i; step=ans=cnt=0; } int findf(int u) { if(u==f[u]) return u; else { f[u]=findf(f[u]); return f[u]; } } void join(int u,int v) { int t1=findf(u);int t2=findf(v); if(t1!=t2) f[t2]=t1; } void add(int u,int v,int id) { edge[cnt].next=head[u];edge[cnt].to=v;edge[cnt].id=id;head[u]=cnt++; edge[cnt].next=head[v];edge[cnt].to=u;edge[cnt].id=id;head[v]=cnt++; } void tarjan(int u,int fa) { low[u]=dfn[u]=++step; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].to;int id=edge[i].id; if(fa==id) continue; if(!dfn[v]) { tarjan(v,id); low[u]=min(low[u],low[v]); if(dfn[u]<low[v]) { ans++; } else { join(u,v); } } else { low[u]=min(low[u],dfn[v]); } } } void bfs(int u) { visit[u]=1;dist[u]=0; queue<int>q; q.push(u); while(!q.empty()) { int x=q.front();q.pop(); for(int i=head[x];i!=-1;i=edge[i].next) { int v=edge[i].to; if(visit[v]) continue; if(findf(x)==findf(v)) { q.push(v);dist[v]=dist[x];visit[v]=1; } else { q.push(v);dist[v]=dist[x]+1;visit[v]=1; } } } } int main() { int x,y; while(scanf("%d%d",&n,&m)&&n&&m) { init(); for(int i=1;i<=m;i++) { scanf("%d%d",&x,&y); add(x,y,i); } tarjan(1,0); bfs(1); int maxx=0;int pos=1; for(int i=1;i<=n;i++) { if(maxx<dist[i]) { maxx=dist[i];pos=i; } } memset(visit,0,sizeof(visit));memset(dist,0,sizeof(dist)); bfs(pos); maxx=0; for(int i=1;i<=n;i++) { if(maxx<dist[i]) maxx=dist[i]; } printf("%d\n",ans-maxx); } }