专题训练之双连通
桥和割点例题+讲解：hihocoder1183 http://hihocoder.com/problemset/problem/1183
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #include<set> 6 using namespace std; 7 const int maxn=1005; 8 const int maxm=200010; 9 struct edge{ 10 int to,nxt; 11 bool cut; 12 }edge[maxm*2]; 13 int head[maxn],tot; 14 int low[maxn],dfn[maxn]; 15 int index,n,bridge; 16 set<int>st; 17 bool cut[maxn]; 18 19 void addedge(int u,int v) 20 { 21 edge[tot].to=v; 22 edge[tot].nxt=head[u]; 23 edge[tot].cut=false; 24 head[u]=tot++; 25 } 26 27 void tarjan(int u,int pre) 28 { 29 low[u]=dfn[u]=++index; 30 int son=0; 31 for ( int i=head[u];i!=-1;i=edge[i].nxt ) { 32 int v=edge[i].to; 33 if ( v==pre ) continue; 34 if ( !dfn[v] ) { 35 son++; 36 tarjan(v,u); 37 low[u]=min(low[u],low[v]); 38 if ( low[v]>dfn[u] ) { 39 bridge++; 40 edge[i].cut=true; 41 edge[i^1].cut=true; 42 } 43 if ( low[v]>=dfn[u] && u!=pre ) { 44 st.insert(u); 45 cut[u]=true; 46 } 47 } 48 else if ( low[u]>dfn[v] ) low[u]=dfn[v]; 49 } 50 if ( u==pre && son>1 ) { 51 cut[u]=true; 52 st.insert(u); 53 } 54 } 55 56 void solve() 57 { 58 memset(low,0,sizeof(low)); 59 memset(dfn,0,sizeof(dfn)); 60 memset(cut,false,sizeof(cut)); 61 index=bridge=0; 62 st.clear(); 63 for ( int i=1;i<=n;i++ ) { 64 if ( !dfn[i] ) tarjan(i,i); 65 } 66 set<int>::iterator it; 67 if ( st.size()==0 ) printf("Null\n"); 68 else { 69 for ( it=st.begin();it!=st.end();it++ ) { 70 if ( it!=st.begin() ) printf(" "); 71 printf("%d",*it); 72 } 73 printf("\n"); 74 } 75 vector<pair<int,int> >ans; 76 for ( int i=1;i<=n;i++ ) { 77 for ( int j=head[i];j!=-1;j=edge[j].nxt ) { 78 if ( edge[j].cut && edge[j].to>i ) ans.push_back(make_pair(i,edge[j].to)); 79 } 80 } 81 sort(ans.begin(),ans.end()); 82 for ( int i=0;i<ans.size();i++ ) { 83 printf("%d %d\n",ans[i].first,ans[i].second); 84 } 85 } 86 87 void init() 88 { 89 tot=0; 90 memset(head,-1,sizeof(head)); 91 } 92 93 int main() 94 { 95 int m,i,j,k,x,y,z; 96 while ( scanf("%d%d",&n,&m)!=EOF ) { 97 init(); 98 while ( m-- ) { 99 scanf("%d%d",&x,&y); 100 addedge(x,y); 101 addedge(y,x); 102 } 103 solve(); 104 } 105 }
边双连通模板
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 using namespace std; 6 const int maxn=1005; 7 const int maxm=200010; 8 struct edge{ 9 int to,nxt; 10 bool cut; 11 }edge[maxm*2]; 12 int head[maxn],tot,n; 13 int index,ebc_cnt,bridge; 14 int dfn[maxn],low[maxn]; 15 16 void addedge(int u,int v) 17 { 18 edge[tot].to=v; 19 edge[tot].nxt=head[u]; 20 edge[tot].cut=false; 21 head[u]=tot++; 22 } 23 24 void tarjan(int u,int pre) 25 { 26 low[u]=dfn[u]=++index; 27 for ( int i=head[u];i!=-1;i=edge[i].nxt ) { 28 int v=edge[i].to; 29 if ( v==pre ) continue; 30 if ( !dfn[v] ) { 31 tarjan(v,u); 32 low[u]=min(low[u],low[v]); 33 if ( low[v]>dfn[u] ) { 34 bridge++; 35 edge[i].cut=true; 36 edge[i^1].cut=true; 37 } 38 } 39 else if ( low[u]>dfn[v] ) low[u]=dfn[v]; 40 } 41 } 42 43 void solve() 44 { 45 memset(low,0,sizeof(low)); 46 memset(dfn,0,sizeof(dfn)); 47 index=bridge=0; 48 for ( int i=1;i<=n;i++ ) { 49 if ( !dfn[i] ) tarjan(i,i); 50 } 51 } 52 53 void init() 54 { 55 tot=0; 56 memset(head,-1,sizeof(head)); 57 }
点双连通模板
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<stack> 5 #include<vector> 6 using namespace std; 7 const int maxn=1005; 8 const int maxm=200010; 9 struct edge{ 10 int to,nxt; 11 }edge[maxm*2]; 12 struct Edge{ 13 int u,v; 14 Edge(int _u=0,int _v=0):u(_u),v(_v) {} 15 }; 16 int dfn[maxn],low[maxn]; 17 int head[maxn],tot; 18 int index,n; 19 bool cut[maxn]; 20 int bcc_cnt,bccno[maxn]; 21 vector<int>bcc[maxn]; 22 stack<Edge>s; 23 24 void addedge(int u,int v) 25 { 26 edge[tot].to=v; 27 edge[tot].nxt=head[u]; 28 head[u]=tot++; 29 } 30 31 void tarjan(int u,int pre) 32 { 33 low[u]=dfn[u]=++index; 34 int son=0; 35 for ( int i=head[u];i!=-1;i=edge[i].nxt ) { 36 int v=edge[i].to; 37 if ( v==pre ) continue; 38 Edge e(u,v); 39 if ( !dfn[v] ) { 40 s.push(e); 41 son++; 42 tarjan(v,u); 43 low[u]=min(low[u],low[v]); 44 if ( low[v]>=dfn[u] ) { 45 cut[u]=true; 46 bcc_cnt++; 47 bcc[bcc_cnt].clear(); 48 for (;;) 49 { 50 Edge x=s.top(); 51 s.pop(); 52 if ( bccno[x.u]!=bcc_cnt ) { 53 bcc[bcc_cnt].push_back(x.u); 54 bccno[x.u]=bcc_cnt; 55 } 56 if ( bccno[x.v]!=bcc_cnt ) { 57 bcc[bcc_cnt].push_back(x.v); 58 bccno[x.v]=bcc_cnt; 59 } 60 if ( x.u==u && x.v==v ) break; 61 } 62 } 63 } 64 else if ( dfn[v]<dfn[u] && v!=pre ) { 65 s.push(e); 66 low[u]=min(low[u],dfn[v]); 67 } 68 } 69 if ( u==pre && son==1 ) cut[u]=false; 70 } 71 72 void solve() 73 { 74 memset(dfn,0,sizeof(dfn)); 75 memset(low,0,sizeof(low)); 76 memset(cut,false,sizeof(cut)); 77 index=bcc_cnt=0; 78 while ( !s.empty() ) s.pop(); 79 for ( int i=1;i<=n;i++ ) { 80 if ( !dfn[i] ) tarjan(i,i); 81 } 82 } 83 84 void init() 85 { 86 tot=0; 87 memset(head,-1,sizeof(head)); 88 }
1.（POJ2117）http://poj.org/problem?id=2117 （求连通块数量）
题意：去掉一个点使得有更多的连通块，求最多有多少连通块
分析：添加数组add_block[]，当u为割点时则add_block[u]++，最后逐一枚举要去掉的点。特别注意对于数根来说add_block[u]=son-1
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #include<set> 6 using namespace std; 7 const int maxn=20005; 8 const int maxm=100010; 9 struct edge{ 10 int to,nxt; 11 bool cut; 12 }edge[maxm*3]; 13 int head[maxn],tot; 14 int low[maxn],dfn[maxn]; 15 int index,n,bridge; 16 int add_block[maxn]; 17 set<int>st; 18 bool cut[maxn]; 19 20 void addedge(int u,int v) 21 { 22 edge[tot].to=v; 23 edge[tot].nxt=head[u]; 24 edge[tot].cut=false; 25 head[u]=tot++; 26 } 27 28 void tarjan(int u,int pre) 29 { 30 low[u]=dfn[u]=++index; 31 int son=0; 32 for ( int i=head[u];i!=-1;i=edge[i].nxt ) { 33 int v=edge[i].to; 34 if ( v==pre ) continue; 35 if ( !dfn[v] ) { 36 son++; 37 tarjan(v,u); 38 low[u]=min(low[u],low[v]); 39 if ( low[v]>dfn[u] ) { 40 bridge++; 41 edge[i].cut=true; 42 edge[i^1].cut=true; 43 } 44 if ( u!=pre && low[v]>=dfn[u] ) { 45 st.insert(u); 46 cut[u]=true; 47 add_block[u]++; 48 } 49 } 50 else if ( low[u]>dfn[v] ) low[u]=dfn[v]; 51 } 52 if ( u==pre && son>1 ) { 53 cut[u]=true; 54 st.insert(u); 55 } 56 if ( u==pre ) add_block[u]=son-1; 57 } 58 59 void solve() 60 { 61 memset(low,0,sizeof(low)); 62 memset(dfn,0,sizeof(dfn)); 63 memset(cut,false,sizeof(cut)); 64 memset(add_block,0,sizeof(add_block)); 65 int cnt,ans; 66 index=bridge=cnt=ans=0; 67 for ( int i=1;i<=n;i++ ) { 68 if ( !dfn[i] ) { 69 tarjan(i,i); 70 cnt++; 71 } 72 } 73 for ( int i=1;i<=n;i++ ) ans=max(ans,cnt+add_block[i]); 74 printf("%d\n",ans); 75 } 76 77 void init() 78 { 79 tot=0; 80 memset(head,-1,sizeof(head)); 81 st.clear(); 82 } 83 84 int main() 85 { 86 int m,i,j,k,x,y,z; 87 while ( scanf("%d%d",&n,&m)!=EOF && (n+m) ) { 88 init(); 89 while ( m-- ) { 90 scanf("%d%d",&x,&y); 91 x++;y++; 92 addedge(x,y); 93 addedge(y,x); 94 } 95 solve(); 96 } 97 }
2.(POJ3117)http://poj.org/problem?id=3177 （构造边双连通）
题意：求添加多少条边后在图中的任意两点都有两条边不重复的路径
分析：边双连通，利用强连通分量中的写法，把每个点对应的缩点后的点标记下来。最后构建新图（即进行缩点，边只存在原先为桥的边）记录入度（或者出度，选择一个即可），最后入度为1的点即为叶子节点，对于一棵树想要使其变成边双连通，所加的边数=（叶子节点的个数+1）/2
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 using namespace std; 6 const int maxn=1005; 7 const int maxm=200010; 8 struct edge{ 9 int to,nxt; 10 bool cut; 11 }edge[maxm*2]; 12 int head[maxn],tot,n; 13 int index,ebc_cnt,bridge,block,top; 14 int dfn[maxn],low[maxn],belong[maxn],stack[maxn],du[maxn]; 15 bool vis[maxn]; 16 17 void addedge(int u,int v) 18 { 19 edge[tot].to=v; 20 edge[tot].nxt=head[u]; 21 edge[tot].cut=false; 22 head[u]=tot++; 23 } 24 25 void tarjan(int u,int pre) 26 { 27 low[u]=dfn[u]=++index; 28 stack[top++]=u; 29 vis[u]=true; 30 for ( int i=head[u];i!=-1;i=edge[i].nxt ) { 31 int v=edge[i].to; 32 if ( v==pre ) continue; 33 if ( !dfn[v] ) { 34 tarjan(v,u); 35 low[u]=min(low[u],low[v]); 36 if ( low[v]>dfn[u] ) { 37 bridge++; 38 edge[i].cut=true; 39 edge[i^1].cut=true; 40 } 41 } 42 else if ( low[u]>dfn[v] && vis[v] ) low[u]=dfn[v]; 43 } 44 if ( low[u]==dfn[u] ) { 45 block++; 46 int v; 47 do 48 { 49 v=stack[--top]; 50 vis[v]=true; 51 belong[v]=block; 52 } 53 while ( v!=u ); 54 } 55 } 56 57 void solve() 58 { 59 memset(low,0,sizeof(low)); 60 memset(dfn,0,sizeof(dfn)); 61 memset(vis,false,sizeof(vis)); 62 index=bridge=block=top=0; 63 for ( int i=1;i<=n;i++ ) { 64 if ( !dfn[i] ) tarjan(i,i); 65 } 66 memset(du,0,sizeof(du)); 67 for ( int i=1;i<=n;i++ ) { 68 for ( int j=head[i];j!=-1;j=edge[j].nxt ) { 69 if ( edge[j].cut ) { 70 du[belong[i]]++; 71 } 72 } 73 } 74 int cnt=0; 75 for ( int i=1;i<=block;i++ ) { 76 if ( du[i]==1 ) cnt++; 77 } 78 printf("%d\n",(cnt+1)/2); 79 } 80 81 void init() 82 { 83 tot=0; 84 memset(head,-1,sizeof(head)); 85 } 86 87 int main() 88 { 89 int m,i,j,k,x,y,z; 90 while ( scanf("%d%d",&n,&m)!=EOF ) { 91 init(); 92 for ( i=1;i<=m;i++ ) { 93 scanf("%d%d",&x,&y); 94 addedge(x,y); 95 addedge(y,x); 96 } 97 solve(); 98 } 99 return 0; 100 }
3.（HDOJ2242）http://acm.hdu.edu.cn/showproblem.php?pid=2242
分析：边双连通+搜索。首先分成3类情况，对于初始时连通块的数量>2或者连通块为1但不存在桥直接输出impossible。而对于连通块本身为2的图来说不用切断如何一条管道即可将整个图分成两部分。最后仅仅需要对连通块数量为1，同时桥的数量>0的图进行考虑即可。给每个点标号标记它们属于第几个边强连通分量，然后构建新图。最后对新图进行搜索，不断更新答案.同时注意存在重flag解决存在重边的问题。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #include<cmath> 6 using namespace std; 7 const int maxn=10005; 8 const int maxm=20010; 9 struct edge{ 10 int to,nxt; 11 bool cut; 12 }edge[maxm*2]; 13 int head[maxn],tot,n; 14 int index,ebc_cnt,bridge,block,top,ans,sum; 15 int dfn[maxn],low[maxn],belong[maxn],stack[maxn],num[maxn],num_[maxn]; 16 bool vis[maxn]; 17 vector<int>G[maxn]; 18 19 void addedge(int u,int v) 20 { 21 edge[tot].to=v; 22 edge[tot].nxt=head[u]; 23 edge[tot].cut=false; 24 head[u]=tot++; 25 } 26 27 void addedge_(int u,int v) 28 { 29 G[u].push_back(v); 30 } 31 32 void tarjan(int u,int pre) 33 { 34 low[u]=dfn[u]=++index; 35 stack[top++]=u; 36 vis[u]=true; 37 bool flag=false; 38 for ( int i=head[u];i!=-1;i=edge[i].nxt ) { 39 int v=edge[i].to; 40 if ( v==pre && !flag ) { 41 flag=true; 42 continue; 43 } 44 if ( !dfn[v] ) { 45 tarjan(v,u); 46 low[u]=min(low[u],low[v]); 47 if ( low[v]>dfn[u] ) { 48 bridge++; 49 edge[i].cut=true; 50 edge[i^1].cut=true; 51 } 52 } 53 else if ( low[u]>dfn[v] && vis[v] ) low[u]=dfn[v]; 54 } 55 if ( low[u]==dfn[u] ) { 56 block++; 57 int v; 58 do 59 { 60 v=stack[--top]; 61 vis[v]=true; 62 belong[v]=block; 63 } 64 while ( v!=u ); 65 } 66 } 67 68 int dfs(int u,int pre) 69 { 70 int now=num_[u]; 71 for ( int i=0;i<G[u].size();i++ ) { 72 int v=G[u][i]; 73 if ( v!=pre ) now+=dfs(v,u); 74 } 75 ans=min(ans,abs(sum-2*now)); 76 return now; 77 } 78 79 void solve() 80 { 81 memset(low,0,sizeof(low)); 82 memset(dfn,0,sizeof(dfn)); 83 memset(vis,false,sizeof(vis)); 84 int cnt,now; 85 index=bridge=block=top=now=cnt=0; 86 for ( int i=1;i<=n;i++ ) { 87 if ( !dfn[i] ) { 88 tarjan(i,i); 89 cnt++; 90 } 91 if ( i==1 ) { 92 for ( int j=1;j<=n;j++ ) { 93 if ( dfn[j] ) now+=num[j]; 94 } 95 } 96 } 97 memset(num_,0,sizeof(num_)); 98 for ( int i=1;i<=n;i++ ) num_[belong[i]]+=num[i]; 99 if ( cnt>2 || bridge==0 ) { 100 printf("impossible\n"); 101 return; 102 } 103 else if ( cnt==2 ) { 104 printf("%d\n",abs(sum-2*now)); 105 return; 106 } 107 for ( int i=1;i<=block;i++ ) G[i].clear(); 108 for ( int i=1;i<=n;i++ ) { 109 for ( int j=head[i];j!=-1;j=edge[j].nxt ) { 110 int v=edge[j].to; 111 if ( edge[j].cut ) { 112 int x=belong[i]; 113 int y=belong[v]; 114 addedge_(x,y); 115 } 116 } 117 } 118 ans=sum; 119 dfs(1,1); 120 printf("%d\n",ans); 121 } 122 123 void init() 124 { 125 tot=0; 126 memset(head,-1,sizeof(head)); 127 } 128 129 int main() 130 { 131 int m,i,j,k,x,y,z; 132 while ( scanf("%d%d",&n,&m)!=EOF ) { 133 init(); 134 sum=0; 135 for ( i=1;i<=n;i++ ) { 136 scanf("%d",&num[i]); 137 sum+=num[i]; 138 } 139 for ( i=1;i<=m;i++ ) { 140 scanf("%d%d",&x,&y); 141 x++;y++; 142 addedge(x,y); 143 addedge(y,x); 144 } 145 solve(); 146 } 147 return 0; 148 }
