之前只是听说过双连通分量,没有自己去写过,这次遇到了果断没有做出来,看题的时候算法是想到了的,奈何实在太挫了,写不出来。之前想用dfs找环再标记缩点,结果wa了,不知道怎么调就放弃了。今天看了一下Tarjan找桥和割点的方法还是比较简单的,当dfn[u]<low[v]时,就表示v顶点不经过父子边能到达的最远的点dfs编号比u小,就说明没有其他路可以到达u,故u->v是桥,然后标记一下,重新建图,求一个直径就好了。
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <cstring> #include <cstdio> #include <stack> #include <queue> #include <vector> using namespace std; #define MAXM 2000005 #define MAXN 200005 struct Edge { int v,next; bool inbridge; }edge[MAXM]; int head[MAXN],e; void addedge(int u,int v) { edge[e].v=v; edge[e].next=head[u]; edge[e].inbridge=false; head[u]=e++; edge[e].v=u; edge[e].next=head[v]; edge[e].inbridge=false; head[v]=e++; } int dfn[MAXN],low[MAXN],sign,bridge; int instack[MAXN],scnt,belong[MAXN]; stack<int> s; vector<int> g[MAXN]; void tarjan(int u,int p) { s.push(u); dfn[u]=low[u]=++sign; bool flag=true; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].v; if(v==p&&flag) { flag=false; continue; } if(!dfn[v]) { tarjan(v,u); if(low[u]>low[v]) low[u]=low[v]; if(dfn[u]<low[v]) { bridge++; edge[i].inbridge=true; edge[i^1].inbridge=true; } } else if(low[u]>dfn[v]) low[u]=dfn[v]; } if(dfn[u]==low[u]) { int i; scnt++; do { i=s.top();s.pop(); instack[i]=false; belong[i]=scnt; } while(i!=u); } } void work() { memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(instack,0,sizeof(instack)); sign=scnt=bridge=0; tarjan(1,0); } struct Node { int v,dis; Node(int a,int b){v=a,dis=b;} }; bool vis[MAXN]; int maxlen; int bfs(int k) { memset(vis,0,sizeof(vis)); queue<Node> q; q.push(Node(k,0)); int len=0,node=1; vis[k]=true; while(!q.empty()) { Node u=q.front(); q.pop(); for(int i=0;i<g[u.v].size();i++) { int v=g[u.v][i]; if(!vis[v]) { vis[v]=true; if(len<u.dis+1) { len=u.dis+1; node=v; } q.push(Node(v,u.dis+1)); } } } maxlen=len; return node; } int main() { //freopen("input.txt","r",stdin); int n,m; while(scanf("%d%d",&n,&m),n+m) { memset(head,-1,sizeof(head)); e=0; for(int i=1;i<=n;i++) g[i].clear(); for(int i=0;i<m;i++) { int u,v; scanf("%d%d",&u,&v); addedge(u,v); } work(); for(int i=1;i<=n;i++) { for(int j=head[i];j!=-1;j=edge[j].next) { if(edge[j].inbridge) { g[belong[i]].push_back(belong[edge[j].v]); } } } int u=bfs(1); int v=bfs(u); printf("%d\n",bridge-maxlen); } }