标签:
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5093 Accepted Submission(s):
1131
//scc代表 双联通分量 写习惯了 顺手就写成scc了也懒得改了
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#define MAXM 2000100
#define MAX 200100
#define INF 0x7fffff
using namespace std;
int n,m,bridge,sum;
int low[MAX],dfn[MAX];
int head[MAX],ans,age;
int sccno[MAX];//代表当前点属于哪个双连通分量
int dfsclock,scccnt;
vector<int>newmap[MAX];//储存缩点后的新图
int instack[MAX];//标记当前点是否入栈
int dis[MAX];//求树的直径时记录路径的长度
int vis[MAX];//求树的直径时标记是否入队列
stack<int>s;
struct node
{
int beg,end,next;
}edge[MAXM];
void init()
{
ans=0;
bridge=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
edge[ans].beg=u;
edge[ans].end=v;
edge[ans].next=head[u];
head[u]=ans++;
}
void getmap()
{
int a,b;
while(m--)
{
scanf("%d%d",&a,&b);
add(a,b);
add(b,a);
}
}
void tarjan(int u,int fa)
{
int v,i;
low[u]=dfn[u]=++dfsclock;
instack[u]=1;
s.push(u);
int flag=1;
for(i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].end;
if(flag&&v==fa)//判重边
{
flag=0;
continue;
}
if(!dfn[v])
{
tarjan(v,u);
low[u]=min(low[u],low[v]);
if(dfn[u]<low[v])
bridge++;
}
else if(instack[v])
low[u]=min(low[u],dfn[v]);
}
if(dfn[u]==low[u])
{
scccnt++;
while(1)
{
v=s.top();
s.pop();
instack[v]=0;
sccno[v]=scccnt;
if(v==u)
break;
}
}
}
void find()
{
int i;
memset(low,0,sizeof(low));
memset(sccno,0,sizeof(sccno));
memset(dfn,0,sizeof(dfn));
memset(instack,0,sizeof(instack));
dfsclock=scccnt=0;
for(i=1;i<=n;i++)
{
if(!dfn[i])
tarjan(i,-1);
}
}
void suodian()
{
int u,v,i,j;
for(i=1;i<=scccnt;i++)
newmap[i].clear();
for(i=0;i<ans;i=i+2)
{
u=sccno[edge[i].beg];
v=sccno[edge[i].end];
if(u!=v)
{
newmap[u].push_back(v);
newmap[v].push_back(u);
}
}
}
void bfs(int beg)
{
queue<int>q;
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));
int i,j;
while(!q.empty())
q.pop();
sum=0;
age=beg;
vis[beg]=1;
q.push(beg);
int u;
while(!q.empty())
{
u=q.front();
q.pop();
for(i=0;i<newmap[u].size();i++)
{
if(!vis[newmap[u][i]])
{
dis[newmap[u][i]]=dis[u]+1;
vis[newmap[u][i]]=1;
q.push(newmap[u][i]);
if(sum<dis[newmap[u][i]])
{
sum=dis[newmap[u][i]];
age=newmap[u][i];
}
}
}
}
}
void solve()
{
int i,j;
bfs(1);
bfs(age);
printf("%d\n",bridge-sum);
//printf("%d\n%d\n",bridge,sum);
}
int main()
{
while(scanf("%d%d",&n,&m),n|m)
{
init();
getmap();
find();
//printf("%d#\n",bridge);
suodian();
solve();
}
return 0;
}
hdoj 4612 Warm up【双连通分量求桥&&缩点建新图求树的直径】
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原文地址:http://www.cnblogs.com/tonghao/p/4874499.html