传送门:
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=737
题目大意:
给出多个连通的无向图(不止一个),求出各个连通图中的桥,并把所有按顺序输出,求出的桥的两个顶点也需按从小到大。
解题思路:
对每个点进行DFS搜索展开,运用定理low(v)>pre(u)求得各个连通图中的所有桥,放入一集合中并对集合排序并且判断重边即可。
最大的坑:格式,每个样例后要输出一空白行,特别是顶点数n=0的特例特判输出是最容易忘记输出空白行!
Code:
/* W w w mm mm 222222222 7777777777777 */
/* W w w w m m m m 222 22 7777 */
/* w w w w m m m m 22 777 */
/* w w w w m m m m 22 77 */
/* w w w w m m m m 222 77 */
/* w w w w m m m m 222 77 */
/* w w w w m m m m 222 77 */
/* w w w w m m m m 222 77 */
/* w w w w m m m m 222 77 */
/* ww ww m mm m 222222222222222 77 */
//#pragma comment(linker, "/STACK:102400000,102400000")
//C++
//int size = 256 << 20; // 256MB
//char *p = (char*)malloc(size) + size;
//__asm__("movl %0, %%esp\n" :: "r"(p));
//G++
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<ctime>
#include<deque>
#include<cmath>
#include<vector>
#include<string>
#include<cctype>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
#define REP(i,s,t) for(int i=(s);i<=(t);i++)
#define REP2(i,t,s) for(int i=(t);i>=s;i--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned long ul;
const int N=100005;
int n;
struct CountBridge
{
vector<int>G[N];
int low[N],pre[N];
bool iscut[N];
int dfs_clock;
vector<pair<int,int> >bri;
void init()
{
REP(i,0,n-1)
{
G[i].clear();
}
bri.clear();
memset(pre,0,sizeof(pre));
memset(iscut,false,sizeof(iscut));
dfs_clock=0;
}
void addedge(int u,int v)
{
G[u].push_back(v);
G[v].push_back(u);
}
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];
if(!pre[v])
{
child++;
int lowv=dfs(v,u);
lowu=min(lowu,lowv);
if(lowv>=pre[u])
{
iscut[u]=true;
if(lowv>pre[u])
{
if(u<v)
{
bri.push_back(make_pair(u,v));
}
else
{
bri.push_back(make_pair(v,u));
}
}
}
}
else if(pre[v]<pre[u]&&v!=fa)
{
lowu=min(lowu,pre[v]);
}
}
if(fa<0&&child==1)
{
iscut[u]=0;
}
low[u]=lowu;
return lowu;
}
int solve()
{
REP(i,0,n-1)
{
if(!pre[i])
{
dfs(i,-1);
}
}
printf("%d critical links\n",bri.size());
sort(bri.begin(),bri.end());
vector<pair<int,int> >::iterator yy,it;
yy=unique(bri.begin(),bri.end());
//for(int i=0; i<bri.size(); i++)
{
//printf("%d - %d\n",bri[i].first,bri[i].second);
}
for(it=bri.begin();it!=bri.end();it++)
{
printf("%d - %d\n",(*it).first,(*it).second);
}
printf("\n");
}
void debug()
{
REP(i,0,n-1)
{
if(G[i].size()==0)
{
continue;
}
printf("%d ",i);
for(int j=0; j<G[i].size(); j++)
{
printf(" %d",G[i][j]);
}
printf("\n");
}
REP(i,0,n-1)
{
printf("low[%d]=%d\n",i,low[i]);
}
printf("dfs_clock=%d\n",dfs_clock);
}
} solver;
int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("test.in","r",stdin);
#endif
while(~scanf("%d",&n))
{
if(n==0)
{
printf("0 critical links\n\n");
continue;
}
solver.init();
int u;
REP(i,1,n)
{
int x;
scanf("%d (%d)",&u,&x);
REP(j,1,x)
{
int v;
scanf("%d",&v);
solver.addedge(u,v);
}
}
solver.solve();
//solver.debug();
}
return 0;
}