标签：i++ space long algorithm tac 多少 存在 ini coder
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 }
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双连通，求最多有多少双连通
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 }
标签：i++ space long algorithm tac 多少 存在 ini coder
原文地址：https://www.cnblogs.com/HDUjackyan/p/8822773.html
