poj 3352 Road Construction 无向图边双连通分量
2017-08-22 21:51
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题目链接:
http://poj.org/problem?id=3352题意
给个无向图判断加多少边能使之变成边双连通图无向图双连通分量求法:浅谈双连通分量、强连通分量
思路
根据low值来判断点是否在一个双连通分量据说是错误的,可能low值不同的也在同一个双连通分量中。所以应该由栈来记录,类似求强连通分量。使整个图变成双连通图需要加的边数为缩点后形成的树中叶子的个数(leaf+1)/2。过程为先找出两个最近公共祖先最远的两个叶子节点,它们之间加一条边,形成环后必是双连通了,重复这个过程。共(leaf+1)/2次。#include<cstdio> // 无向图加多少条边后:变成一个双连通图 #include<queue> #include<iostream> #include<vector> #include<map> #include<cstring> #include<string> #include<set> #include<stack> #include<algorithm> #define cle(a) memset(a,0,sizeof(a)) #define inf(a) memset(a,0x3f,sizeof(a)) #define ll long long #define Rep(i,a,n) for(int i=a;i<=n;i++) using namespace std; const int INF = ( 2e9 ) + 2; const ll maxn = 1e3+10; int dg[maxn],dfn[maxn],low[maxn],vis[maxn],head[maxn]; int scc[maxn]; stack<int> s; int num; struct edge { int to,next; } e[maxn*3]; int tot,time; void addedge(int u,int v) { e[tot].to=v; e[tot].next=head[u]; head[u]=tot++; e[tot].to=u; e[tot].next=head[v]; head[v]=tot++; } void tarjan(int u,int fa) { low[u]=dfn[u]=++time; vis[u]=1; s.push(u); for(int i=head[u]; i!=-1; i=e[i].next) { int v=e[i].to; if(v==fa)continue; if(!dfn[v]) { tarjan(v,u); low[u]=min(low[u],low[v]); } d656 else if(vis[v]) low[u]=min(low[u],low[v]); } if(low[u]==dfn[u]) { int x; num++; do { x=s.top(); s.pop(); vis[x]=0; scc[x]=num; } while(x!=u); } } void init() { memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(head,-1,sizeof(head)); memset(dg,0,sizeof(dg)); memset(scc,0,sizeof(scc)); time=tot=num=0; while(!s.empty())s.pop(); } int main() { int n,m; while(~scanf("%d%d",&n,&m)) { init(); for(int i=0; i<m; i++) { int u,v; scanf("%d%d",&u,&v); addedge(u,v); } tarjan(1,-1); for(int i=1; i<=n; i++) for(int j=head[i]; j!=-1; j=e[j].next) { int v=e[j].to; if(scc[v]!=scc[i]) { dg[scc[i]]++; } } int cnt=0; for(int i=1; i<=n; i++) if(dg[i]==1)cnt++; printf("%d\n",(cnt+1)/2); } }
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