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 }
桥和割点模板
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 }
POJ2117
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 }
POJ3117
原文地址：https://www.cnblogs.com/HDUjackyan/p/8822773.html
时间： 2024-11-03 23:25:36